MAT137
1
1 SET NOTATION
Set Notation
Elements and Sets
Typically, we use lowercase letters such as x, y to denote elements (things).
We use uppercase letters such as A, B to denote sets (a bag of things).
Sets are made up of elements.
If an element x is inside set A, we write x 2 A.
Set Compliment
Ac stands for the set complement of A.
It contains ’everything’ except for the things in A.
Set Containment
We use the symbol ⇢ or ✓ to denote contained in.
For example, A ⇢ B means set A is contained in set B.
This means every element x in A, x is also in B.
Then we arrive at the property of transitivity for set containment:
If A ⇢ B and B ⇢ C, Then A ⇢ C
If we interpret set A, B, C as bags, then the statement basically says:
If bag A is inside bag B, but bag B is inside bag C,
then bag A must be inside bag C.
Set Equality
Two sets are equal only when A ⇢ B and B ⇢ A.
Union and Intersection
We use the symbol [ to denote union of sets.
We say an element x 2 A [ B only when x 2 A or x 2 B.
We use the symbol \ to denote intersection of sets.
We say an element x 2 A \ B only when x 2 A and x 2 B.
Distributivity and DeMorgan’s Laws
A [ (B \ C) = (A [ B) \ (A [ C)
(A [ B)c = Ac \ B c
Author: Xiao
A \ (B [ C) = (A \ B) [ (A \ C)
(A \ B)c = Ac [ B c
Page 1
MAT137
2
2 PROOF TECHNIQUES
Proof Techniques
Instantiation
Want to prove a statement P is true for many things (everything in set S).
Example statement:
8x 2 S, P is true. (where S is a set, 8 means for every)
1. Pick x 2 S randomly, or ’arbitrarily’.
Typically, things in S will have some common properties.
2. Use the properties to prove P is true for this specific x you picked.
3. Conclude P is true for all things in S.
Contradiction
Want to prove a statement P is true.
Instead of proving P directly:
1. Suppose P is false.
2. After some logical reasoning, deduce an impossible statement.
Example: 1=0
3. Conclude P is true.
Implication
Want to prove the statement: If A is true, then B is true. ( A ) B )
1. Suppose A is true.
2. Using the assumption, prove B is true.
3. Conclude A ) B is true.
Equivalent
Want to prove: A if and only if B. ( A () B)
1. Prove ( A ) B ) is true.
2. Prove ( B ) A ) is true.
3. Conclude A () B is true.
Author: Xiao
Page 2
MAT137
2 PROOF TECHNIQUES
Existence
Want to prove something exist, which makes the statement P to be true.
Example statement:
9x 2 S, P is true. (where S is a set, 9 means there exists at least 1)
1. Give one example of x 2 S.
Typically, x will use quantities that are from former parts of the proof.
x can also be taken directly as a element in S, such as x = 3 2 S.
2. Show that P is true for the specific x you have chosen.
(Thus, we need choose x so that we can show P is true.)
3. Conclude x does exist, which makes P true.
Set Containment
Want to prove a set S1 is contained inside S2 : S1 ⇢ S2 .
Method 1: Instantiation.
1. Pick x 2 S1 randomly, or ’arbitrarily’.
2. Using the common properties of S1 , prove x 2 S2 .
3. Conclude S1 ⇢ S2 .
Method 2: Use contradiction.
1. Suppose S1 6⇢ S2 .
2. Immediately conclude 9x 2 S1 , x 62 S2 .
3. After some logical reasoning, deduce an impossible statement.
4. Conclude S1 ⇢ S2 .
Set Equality
Want to prove 2 sets are equal: S1 = S2 .
1. Prove S1 ⇢ S2 .
2. Prove S2 ⇢ S1 .
3. Conclude S1 = S2 .
Author: Xiao
Page 3
MAT137
2 PROOF TECHNIQUES
Negation
Want to prove a statement Q is not true.
Type 1: Statements with quantifiers: for every 8 and exists 9
statement Q = 8x 2 S1 , 9y 2 S2 , P is true. (where P is another statement.)
1. Write the word not in front of the statement Q with a bracket.
Example: not (Q) = not (8x 2 S1 , 9y 2 S2 , P is true.)
2. Bring the word not toward the right, passing the quantifiers 8, 9.
Each time not pass across 8, change 8 to 9.
Similarly, each time not pass across 9, change 9 to 8.
Example: not (Q) = 9x 2 S1 , not (9y 2 S2 , P is true.)
not (Q) = 9x 2 S1 , 8y 2 S2 , not (P is true.)
not (Q) = 9x 2 S1 , 8y 2 S2 , P is false.
3. Prove the simplified statement of not (Q) attained at the end.
Type 2: Statements with implication: statement Q = (A ) B)
1. Write the word not in front of the statement Q with a bracket.
Example: not (Q) = not(A ) B).
2. Logically, the statement A ) B is equivalent to: not (A) or B.
Use De Morgan’s Law and⇣ bring the word
⌘ not into the bracket.
Example: not (Q) = not not (A) or B
⇣
⌘
not (Q) = not not (A) and not (B)
not (Q) = (A is true) and B is false.
3. Prove the simplified statement of not (Q) attained at the end.
Contrapositive
Want to prove the implication statement: A ) B
Instead of proving the implication directly:
1. We can choose to prove the equivalent statement: not (B) ) not (A).
2. Suppose not (B) is true.
3. Prove not (A) is true.
4. Conclude A ) B is true.
Author: Xiao
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MAT137
2 PROOF TECHNIQUES
Proof By Cases
Split into cases, and prove the statement in both (all) cases.
Type 1: Want to prove the statement: (P or Q) is true.
1. Consider the following two cases.
Case 1: P is true.
Case 2: P is false.
(We need prove (P or Q) is true in both cases.)
2. In case 1:
Conclude (P or Q) is true.
In case 2:
Use the new assumption P is false,
prove Q is true.
Conclude (P or Q) is true.
3. Conclude (P or Q) is true in both cases.
Important Note:
Alternatively, we can do the reverse, and split into the following 2 cases:
Case 1: Q is true.
Case 2: Q is false.
Sometimes P and Q are very di↵erent statements, and only one of the
above 2 case splits can lead to a proof.
Type 2: We know (P or Q) is true. Want to prove statement R is true.
1. Consider the following two cases.
Case 1: P is true.
Case 2: Q is true.
(We need prove R is true in both cases.)
2. In case 1:
Use the new assumption P is true,
prove R is true.
(Do not use Q is true.)
In case 2:
Use the new assumption Q is true,
prove R is true.
(Do not use P is true.)
3. Conclude R is true in both cases.
Induction
Want to prove: 8n 2 N, P (n) is true. (P involves the natural number n)
1. Prove P (1) is true.
2. Suppose P (n) is true.
Prove P (n + 1) is true. (Replace all n in statement P with n + 1)
3. Conclude 8n 2 N, P (n) is true.
Author: Xiao
Page 5
MAT137
2 PROOF TECHNIQUES
Question 1
Prove the following statements are Equivalent:
1. A [ B = B
2. A ⇢ B
3. A \ B = A
4. B c ⇢ Ac
Strategy:
To prove several statements are equivalent,
we typically prove a loop of implications.
In this case, let’s try to prove 1 ) 2, 2 ) 3, 3 ) 4, 4 ) 1.
In general, we can take any order of number we like.
Proof Part 1: Prove 1 ) 2.
This means we need to prove the statement: A [ B = B ) A ⇢ B.
In proving an Implication, the first step is to suppose the left side is true.
So we write in the first line of the proof:
Suppose
.
Next, the statement we need to prove changes to: A ⇢ B.
In proving a set containment, we first take a random element on the left.
So we write in the second line of the proof to be:
Let x 2
be chosen randomly.
Now the statement we need to prove changes to: x 2
.
We need to use our assumption in step 1 to arrive at this conclusion.
Think: Since x 2 A, what is its relationship with A [ B in the assumption.
Since x 2 A,
Since A [ B = B,
Thus x 2
.
Thus 1 ) 2 is true.
Author: Xiao
.
.
Page 6
MAT137
2 PROOF TECHNIQUES
Proof Part 2: Prove 2 ) 3.
The statement we need to prove is: A ⇢ B ) A \ B = A.
Since it’s an implication, we need to:
Suppose
.
Next the statement we need to prove changes to A \ B = A.
Note that this is a Set equality.
We need to prove 2 statements: A \ B ⇢ A and A ⇢ A \ B.
First, we prove: A \ B ⇢ A.
Let x 2
be chosen randomly.
The statement we need to prove changes to: x 2
Using the definition of intersection, we have x 2
.
Thus x 2
.
and x 2
Next, we prove A ⇢ A \ B.
Let x 2
be chosen randomly.
The statement we need to prove changes to: x 2 A \ B.
By the definition of intersection, we need to prove: x 2
By the previous assumption, we already know x 2
.
By the first assumption
, we also have x 2
Thus x 2 A \ B.
.
and x 2
.
.
Thus A \ B = A.
Thus 2 ) 3 is true.
Author: Xiao
Page 7
MAT137
2 PROOF TECHNIQUES
Proof Part 3: Prove 3 ) 4.
The statement we need to prove is: A \ B = A ) B c ⇢ Ac .
Since it’s an implication, we need to:
Suppose
.
Next the statement we need to prove changes to B c ⇢ Ac .
Note that this is a Set containment.
Let x 2
be chosen randomly.
By definition of complement, x 62
. (62 means not in)
The statement we need to prove changes to: x 2
By definition of complement, we need to prove: x 62
We use proof by contradiction.
Suppose x 2
.
Using the first assumption, we have x 2
.
Using the definition of intersection, we have x 2
This is a contradiction, as we already know x 62
Thus we conclude x 62
Thus 3 ) 4 is true.
Author: Xiao
.
.
and x 2
.
.
.
Page 8
MAT137
2 PROOF TECHNIQUES
Proof Part 4: Prove 4 ) 1.
The statement we need to prove is: B c ⇢ Ac ) A [ B = B.
Since it’s an implication, we need to:
Suppose
.
Next the statement we need to prove changes to A [ B = B.
This is again a Set equality.
First, we prove A [ B ⇢ B.
Let x 2 A [ B be chosen randomly.
The statement we need to prove changes to: x 2
.
Note that as x 2 A [ B, then it is a possibility that x 2 A or x 2 B.
Thus to continue, we need to split the proof into 2 cases.
Case 1: x 2 A
Case 2: x 2 B.
(We need prove x 2
is true in both cases.)
We use a little trick:
Thus x 2
.
As x 2 A, we have x 62 Ac .
Using the first assumption
,
and x 62 Ac , we have x 62
.
Thus x 2
.
Thus in both cases, x 2
.
Next, we prove B ⇢ A [ B.
Let x 2
be chosen randomly.
The statement we need to prove changes to: x 2 A [ B .
By definition of union, we need to prove: x 2
By the previous assumption, we have x 2
Thus x 2
.
.
or x 2
.
Thus A [ B = B.
Thus 4 ) 1 is true.
Author: Xiao
Page 9
MAT137
2 PROOF TECHNIQUES
Question 2
Use induction to prove the following statements P (n) to be true for all n 2 N:
1. P (n) :
n
X
k = 1 + 2 + 3 + ... + n =
k=1
2. P (n) :
n
X
n(n + 1)
2
k 2 = 12 + 22 + 32 + ... + n2 =
k=1
n
X
n(n + 1)(2n + 1)
6
n2 (n + 1)2
3. P (n) :
k = 1 + 2 + 3 + ... + n =
4
k=1
Author: Xiao
3
3
3
3
3
Page 10
MAT137
3
✏
3 ✏
LIMIT DEFINITION
Limit Definition
The standard limit is as follows:
lim f (x) = L
x!a
However, what exactly do these symbols mean?
Intuitively, it is understood as follows:
We think of a function as an arrow, taking one value a in the domain X,
and ’shoot’ it to the value f (a) of the codomain Y.
f :X!Y
a 7! f (a)
(In this course, X and Y will be both R, the set of all real numbers.)
We would like to ask:
If instead of starting at a, we are at the point x that is really close to a (in X),
can we make sure that f (x) is really close to f (a) as well (in Y )?
This seems like a reasonable requirement, and is the intuition behind limit.
How can we translate this sentence into mathematical language?
To keep things simple, let’s say the target is L = f (a), and we want to
get f (x) to be close to L. It would be great if f (x) hit exactly L, however,
that is too high of a requirement. Instead, we create an error range around
L. We require f (x) to hit inside this error range. We call the radius of this
error range to be ✏.
After picking this error range, we need to control the values of x we pick.
If x is far from a, we shouldn’t expect f (x) to be in the error range.
So we give a control range centered at a for the values of x we can pick.
We call the radius of this control range to be .
What should the control range radius satisfy?
We should be able to guarantee that for any value of x taken from the
control range near a, f (x) will be inside the error range, near L.
i.e. We can achieve our goal of f (x) being close to L, by concentrating
the value of x around a, to be inside the control range with radius .
Author: Xiao
Page 11
MAT137
3 ✏
LIMIT DEFINITION
Now to put it into mathematical language:
(Recall that the distance between any 2 points a and b, in R is d(a, b) = |b a|)
For any ✏ > 0 as the error radius, there will exist > 0 as the control
radius, such that for all x inside the control radius (which is |x a| < ),
we can guarantee that f (x) is in the error range (which is |f (x) f (a)| < ✏).
In symbols:
lim f (x) = L
x!a
8✏ > 0, 9 > 0, if 0 < |x
a| <
) then |f (x)
L| < ✏
Remember that the value a is a fixed point given to us from the start, and x
is a variable value that can change inside the control range.
So:
|x
a| <
0 < |x
,a
|f (x)
< x < a+
L| < ✏ , L
✏ < f (x) < L + ✏
a| is just a detail: for limits, we do not consider the possibility of x = a.
Using a similar idea, we can say the limit at infinity is done by controlling
x to be very large. So instead of concentrating x inside a control range with
radius , we concentrate x ”at infinity”, by requiring x > N , where N is
chosen to be large.
lim f (x) = L
x!1
8✏ > 0, 9N, if x > N ) then |f (x)
L| < ✏
What about limit at a does not exist?
It means no matter what value of L (for all L), lim f (x) 6= L.
x!a
Negating the statement would interchange 8 and 9, negating the implication
would mean there is a counter-example to the implication, namely, there
is an x in the control range , with f (x) outside the error range ✏.
8L, 9✏ > 0, 8 > 0, 9 x , with 0 < |x
a| <
and |f (x)
L|
✏
One type of limit not existing is f (x) going to infinity. In this case, we do
have a goal for the values of f (x), which is to be very large (larger than a
large number M ), as long as x is concentrated closed enough to the point a.
lim f (x) = +1
x!a
8M > 0, 9 > 0, if 0 < |x
Author: Xiao
a| <
) then f (x) > M
Page 12
MAT137
4
✏
4 ✏
PROOF OF LIMIT DEFINITION
Proof of Limit Definition
The standard statement we need to prove is:
lim f (x) = L
x!a
8✏ > 0, 9 > 0, s.t. if 0 < |x
a| <
) then |f (x)
L| < ✏
We would need to follow the previous proof techniques:
1. Pick a random ✏ > 0.
2. Find a value of which would make the implication to be true.
( can depend on ✏ that was chosen in step 1.)
3. Proving the implication, first assume 0 < |x
4. Attempt to prove the statement |f (x)
a| <
L| < ✏
The problem is step 2, on how we can obtain this .
Thus, we first start rough work as follows:
1. Write the quantity |f (x)
L|
2. Attemp to factor this quantity into two pieces: |x a| · |S|
where S is the extra stu↵ left over from the factorization.
(Note: For any 2 numbers a and b, |a · b| = |a| · |b|.)
3. Hand pick a number k > 0.
Usually pick k = 1, but sometimes need to pick smaller.
4. Write a
k <x<a+k
5. Use step 4 and some algebra to establish the bound on |S|.
Attain m < |S| < M , where m and M are positive with no x involved.
When M does not exist, pick a smaller k from step 3 and repeat.
6. Write |x
a| · |S| < |x
a| · M < · M < ✏
✏
}, so we have both:
M
✏
|x a| <  k and |x a| < 
.
M
Using this , we would write the formal proof.
7. Conclude we pick
Author: Xiao
= min{k,
Page 13
MAT137
4 ✏
PROOF OF LIMIT DEFINITION
Let’s try an example:
Prove
lim x2 = 9
x!3
8✏ > 0, 9 > 0, s.t. if 0 < |x
1. Write the quantity |f (x)
|x2 9|
3| <
L|
) then |x2
9| < ✏
2. Attemp to factor this quantity into two pieces:
|(x 3)(x + 3)| = |x 3| · |x + 3| = |x 3| · |S|
3. Hand pick a number k = 1.
4. Write 3
1<x<3+1)2<x<4
5. Use step 4 and algebra to establish the bound on |S|.
5<x+3<7
5 < |x + 3| < 7
6. Write |x
3| · 7 < · 7 < ✏
✏
7. Conclude we pick = min{1, }, so we have both:
7
✏
|x 3| <  1 and |x 3| <  .
7
Formal Proof :
Let ✏ > 0 be chosen randomly.
✏
Pick = min{1, }.
7
✏
Then, we have  1 and  .
7
Assume 0 < |x a| < .
(Copy the work of step 2, 3, 4, 5)
|f (x) L| = |x2 9| = |x 3| · |x + 3|
Since |x 3| <  1
)
2<x<4
5<x+3<7
5 < |x + 3| < 7
Then |x 3| · |x + 3| < |x 3| · 7
Since |x 3| <
)
|x 3| · 7 < · 7
✏
✏
Since 
)
·7 ·7=✏
7
7
✏
Thus |x2 9| = |x 3| · |x + 3| < |x 3| · 7 < · 7  · 7 = ✏.
7
Author: Xiao
3| · |x + 3| < |x
Page 14
MAT137
4 ✏
PROOF OF LIMIT DEFINITION
Question 3
Prove the following limits.
1.
lim x2 = 4
x!2
8✏ > 0, 9 > 0, s.t. if 0 < |x
Strategy:
Start with the quantity |f (x)
2| <
L| and factor out |x
2.
lim
x!9
p
b) = a2
3.
lim
x!3
4.
p
4| < ✏
2|
x=3
8✏ > 0, 9 > 0, s.t. if 0 < |x
Strategy:
Use the formula (a + b)(a
) then |x2
9| <
p
) then | x
3| < ✏
b2 to eliminate square root.
x+1=2
1
=1
x!1 x
lim
Strategy:
Notice that k = 1 does not provide a bound for the quantity we need,
so we need to pick a smaller k.
5.
1
1
=
x!1 x2 + 1
2
lim
6.
lim |1
x!2
3x| = 5
Strategy:
In our rough work, we may assume x is be concentrated close enough
to 2 (approximately 2) to determine whether the quantity 1 3x is
positive or negative. However, we need to make this precise in our
formal proof (with the value of k).
Author: Xiao
Page 15
MAT137
4 ✏
PROOF OF LIMIT DEFINITION
Question 4
Given L as a constant, suppose
lim f (x) = L
x!0
Prove
lim f (2x) = L
x!0
Strategy:
We need to use our assumption to prove the statement.
This means we need to write the definition of our assumption and
the statement which we need to prove.
We need to use new names for variables in our assumption.
So, we suppose
8✏2 > 0, 9
2
> 0, s.t. (for all x2 ), if 0 < |x2 0| <
2
) then |f (x2 ) L| < ✏2
We need to prove
8✏ > 0, 9 > 0, s.t. (for all x), if 0 < |x
0| <
) then |f (2x)
L| < ✏
Notice that in the second statement, we have the quantity |f (2x)
L|.
Author: Xiao
Page 16
MAT137
4 ✏
PROOF OF LIMIT DEFINITION
Question 5
Part 1:
Suppose we know the following about the function f (x):
if |x
2| < 1 ) then |f (x)
5| < 2
Which of the following statements are necessarily true?
A statement below is necessarily true means:
We can prove that statement given below using the assumption above.
A statement below is not necessarily true (considered as false) means:
We can give a counter-example: a function f (x) which satisfy our assumption above, and does not satisfy that statement given below.)
1. if |x
2| < 1 ) then |f (x)
2. if |x
2| < 0.5 ) then |f (x)
3. if |x
2| < 1 ) then |f (x)
4. if |x
2| < 0.5 ) then |f (x)
5. There exists
Author: Xiao
5| < 3
5| < 2
5| < 1
5| < 1
> 0 such that if |x
2| <
) then |f (x)
5| < 1
Page 17
MAT137
4 ✏
PROOF OF LIMIT DEFINITION
Part 2:
We still have the assumption,
if |x
2| < 1 ) then |f (x)
5| < 2
Now we suppose (moreover that)
lim f (x) = L
x!2
This means, we know the following is true:
8✏ > 0, 9 > 0, s.t. if 0 < |x
2| <
) then |f (x)
L| < ✏
Question:
1. Can L = 2?
(L can equal 2 means that we can give an example f (x) with L = 2.
L can not equal 2 means that we can prove using our assumption
that L can not (can never) equal 2.)
2. Can L = 2.5?
3. Can L = 8? Make a preliminary conclusion.
4. Can L = 5?
5. Can L = 4?
6. Can L = 3? Make a final conclusion.
Author: Xiao
Page 18
MAT137
5
5 LIMIT AND CONTINUITY
Limit and Continuity
Two Sided Limits
The limit at a point a of the function f (x) is equal to the number L:
1. When you let x approach a from the left side of a, the graph of f (x)
gets close to the number L.
2. When you let x approach a from the right side of a, the graph of f (x)
gets close to the number L.
3. The actual value of f (a) does NOT matter at all.
lim f (x) = L
x!a
One Sided Limits
We only let x approach a from the left side of a, and the graph of f (x) get
close to the number L1 :
lim f (x) = L1
x!a
The minus sign next to a indicates that we are approaching from the left.
Similarly, we can approach a from the right to attain another limit L2 :
lim f (x) = L2
x!a+
The plus sign indicates that we are approaching from the right.
Only when the one sided limits exist, and the numbers L1 = L2 , we say
that the limit of f (x) at a (from both sides) exist:
lim f (x) = lim f (x) = L1 = lim+ f (x) = L2
x!a
x!a
x!a
Continuity
Intuitively, f (x) is continuous in an interval if the graph is in one piece.
f (x) is continuous at the point a if the limit at a is equal to f (a):
lim f (x) = f (a)
x!a
f (x) is continuous in an interval means f (x) is continuous at every point
in the interval. Most functions are continuous, most limits can be obtained
by putting x = a in the formula of f (x).
Every standard function we know are continuous in their domains.
Any function transformations (such as +, , ⇥, /, composition) of
continuous functions, are also continuous. (except division by 0)
Author: Xiao
Page 19
MAT137
5 LIMIT AND CONTINUITY
Limits of x approaching infinity
We allow a to be infinite as well. This is understood intuitively as:
As x becomes larger and larger, f (x) gets closer to a number L.
lim f (x) = L
x!1
We can also have a being minus infinity:
lim f (x) = L
x! 1
Limit Properties
First, we need (to know) the following limits to exist,
lim f (x) = L,
x!a
lim g(x) = M
x!a
Then we can attain
lim [f (x) + g(x)] = L + M
x!a
lim [f (x)
x!a
g(x)] = L
M
lim c · f (x) = c · L
x!a
lim [f (x) · g(x)] = L · M
x!a
f (x)
L
=
if M 6= 0
x!a g(x)
M
If f (x) is continuous, then we can move the limit inside f :
lim
lim f (g(x)) = f [lim g(x)]
x!a
x!a
In fact, we only need to assume f (x) is continuous at lim g(x).
x!a
Infinite Limits
We also allow one sided limits to be infinite:
When you let x approach a from one side of a, the graph of f (x) gets larger
and larger. We write:
lim f (x) = 1,
x!a
lim f (x) = 1
x!a+
We can also have limits being minus infinity as well, where the graph goes
lower and lower.
lim f (x) =
x!a
1,
lim f (x) =
x!a+
1
If the limit is infinity, it is regarded as the limit does not exist.
So we can not use the limit properties above.
Author: Xiao
Page 20
MAT137
5 LIMIT AND CONTINUITY
Verticle Asymptotes
When f (x) is given by a fraction, we always require that the denominator is
not zero.
For example:
1
f (x) =
x 1
Clearly, x = 1 is not allowed. As x gets close to 1, or any other number with
division by 0, f (x) will usually become infinite.
This is called a verticle asymptote at the point x = 1.
Because x = 1 is not allowed, the graph can not cross the verticle asymptote.
In terms of limits, we write
lim
x!1
1
x
1
= ±1
where
lim
x!1
1
x
1
=
1
and
lim+
x!1
1
x
1
= +1
1
Note that the quantity does not exist, and is not equal to 1.
0
Only when we take the limit, we can say it is either +1 or 1.
Horizontal Asymptotes
As x approaches ±1, sometimes f (x) will attain a limit.
This is called the horizontal asymptote.
The graph is allowed to cross the horizontal asymptote.
For example:
1
f (x) =
x 1
As x becomes large in either positive or negative direction, the bottom is
1
large. So it can be thought of as
±1
lim
x!1
1
x
1
⇡
1
=0
1
lim
x! 1
1
x
1
⇡
1
=0
1
1
is ’equal’ to 0, since that is the unique answer
±1
we obtain after taking the limit.
In this case, the value of
Author: Xiao
Page 21
MAT137
5 LIMIT AND CONTINUITY
Question 6
Evaluate lim (x2 + 2x ).
x!1
Question 7
Find conditions on A and B so that
8
>
<Ax B x  1
f (x) = 3x
1<x<2
>
: 2
Bx
A 2x
is continuous at x = 1 and discontinuous at x = 2.
Question 8
Are the following statements true or false? Justify your answers.
(Prove the statement is true, or give a counterexample if it is false.)
1. Suppose lim [f (x) + g(x)] exists and lim f (x) exists,
x!a
x!a
then lim g(x) also exists.
x!a
2. Suppose lim [f (x) + g(x)] exists but lim f (x) does not exist,
x!a
x!a
then lim g(x) also does not exist.
x!a
3. Suppose lim f (x) exists but lim g(x) does not exist,
x!a
x!a
then lim [f (x) + g(x)] also does not exist.
x!a
4. Suppose lim f (x) does not exist and lim g(x) does not exist,
x!a
x!a
then lim [f (x) + g(x)] also does not exist.
x!a
5. Suppose lim f (x) = L 6= 0 and lim [f (x)g(x)] = 1,
x!a
x!a
then lim g(x) also exists.
x!a
6. Suppose lim f (x) = 0 and lim [f (x)g(x)] = 1,
x!a
x!a
then lim g(x) does not exist.
x!a
Author: Xiao
Page 22
MAT137
5 LIMIT AND CONTINUITY
Question 9
Prove the ”Continuity law for composition”:
If f (x) is continuous at lim g(x), then we can move limit inside:
x!a
lim f (g(x)) = f [lim g(x)]
x!a
x!a
Deduce this implies:
Composition of continuous functions is continuous.
Author: Xiao
Page 23
MAT137
6
6 LIMIT EVALUATION TECHNIQUES
Limit Evaluation Techniques
Rational Functions
Rational functions is a polynomial dividing a polynomial.
Try to factor the numerator and the denominator. When cancellation occurs,
one can try to plug in x = a to attain limit.
f (x) =
x2 + 3x
x(x + 3)
x
=
=
x2 x 12
(x + 3)(x 4)
x 4
) lim f (x) =
x! 3
3
3
4
=
Rational Functions with x Approaching Infinity
Only the term with the highest exponent in the numerator and denominator
matters when computing limits.
lim f (x) =
x!1
24x6 + 1
24x6
⇡
= 12
2x6 + 30x5 + 1
2x6
) lim f (x) = 12
x!1
To be precise, we divide by the highest exponent on the numerator and
denominator. In this example, x6 is the highest exponent.
6
lim f (x) =
x!1
6
(24x + 1)/x
=
(2x6 + 30x5 + 1)/x6
24 +
1
x6
30
1
2+
+ 6
x
x
=
24 + 0
24
=
= 12
2+0+0
2
In general,
am xm + ... + a0
bn xn + ... + b0
with degree m in the numerator and degree n in the denominator.
The highest exponent of a polynomial is called the degree.
f (x) =
1. m < n
The denominator grows faster, the limit is 0 as horizontal asymptote.
2. m = n
The numerator and denominator grows at the same pace.
am
The limit is f (x) ⇡
as horizontal asymptote.
bn
3. m > n
The numerator grows faster, and the limit is ±1.
am x m
The sign depends on f (x) ⇡
bn x n
Author: Xiao
Page 24
3
7
MAT137
6 LIMIT EVALUATION TECHNIQUES
Square Roots
Try to use the relation a2 b2 = (a + b)(a b). Set a to be the square root,
then we multiply and divide by the other factor, and try to simplify.
This is called multiply by conjugate.
p
p
p
4x + 1 3
4x + 1 3
4x + 1 + 3
4x + 1 32
p
f (x) =
=
·p
=
x 2
x 2
4x + 1 + 3
(x 2)( 4x + 1 + 3)
4(x 2)
4
4
2
p
=p
) lim f (x) = =
x!2
6
3
(x 2)( 4x + 1 + 3)
4x + 1 + 3
Small Angle Approximation
If the limit is given as x ! 0, then we can make the substitution
=
sin(x)
=1
x!0
x
sin(x) ⇡ x ⇡ tan(x)
i.e. lim
tan(x)
=1
x!0
x
lim
Note: This also works for arcsin(x) and arctan(x).
Absolute Values
When absolute value is in f (x), approach from the left side and right side
separately. Determine whether the expression inside the absolute value is
positive or negative, and remove the absolute value sign.
lim+ (
x!0
1
x
1
1
)=
|x|
x
1
=0
x
lim (
x!0
1
x
1
1
)=
|x|
x
1
2
= =
x
x
1
Squeeze Theorem
To attain lim f (x), can try to find g(x) and h(x) so that:
x!a
1. g(x)  f (x)  h(x) near the point a.
2. lim g(x) = L = lim h(x)
x!a
x!a
Then we conclude lim f (x) = L.
x!a
In practice, we want to show lim f (x) = 0. Start at |f (x)|, create a chain
x!a
of inequality, simplify |f (x)| to attain |g(x)|, which has limit 0 :
|f (x)|  ...  |g(x)| ! 0
i.e. lim |g(x)| = 0
x!a
Then we conclude lim f (x) = 0.
x!a
Author: Xiao
Page 25
MAT137
6 LIMIT EVALUATION TECHNIQUES
Question 10
x2
1. lim
x!3 x
h3
h!2 h3
2. lim
2
x!3 x
3. lim
9
3
13. lim
5h2 + 3h + 6
h2 3h + 2
14. lim
x
3
2 x
x!3 (x
3)2
p
x+1 1
5. lim
x!0
x
4. lim
p
y!1
y+4
p
x!0
17. lim e1/x
x!0
t
t!0 sin(2t)
19. lim
sin(2x)
x!0 sin(3x)
20. lim
sin(2z 2 )
z!0 cos(3z)sin2 (5z)
21. lim
tan(x 3)
x!3
2x 6
22. lim
2ex
x!0 sin(2ex )
23. lim x2 sin
1
x
24. lim 2x sin
1
x
8. lim
9. lim
10. lim
11. lim
12. lim
1
t!0
cos(3t)
t2
y
16. lim+ e1/x
18. lim
7. lim
4y + 1
1
sin(1 cos(x))
x!0
xtan(⇡x)
r
1
u+2
15. lim
(
2)
u!2 2
u
u 1
sin(x)
x!1
x
6. lim
p
e3x 2e
x!1 4e3x + e
x!
3x
3x
e3x 2e
1 4e3x + e
3x
3x
3x 5x
x!1 32x + 24x
x!
3x 5x
1 32x + 24x
x! 6
2x + 12
|x + 6|
x!0
x!0
cos
1
x
Question 11
Define the function at x = 5 so that f (x) becomes continuous.
p
2x 1 3
f (x) =
x 5
Author: Xiao
Page 26
MAT137
7
7 TYPES OF DISCONTINUITY
Types of Discontinuity
Recall, f (x) is continuous at point a if
lim f (x) = f (a)
x!a
Note that this equality contains 2 independent requirements:
1. The limit lim f (x) = L would exist (as a 2-sided limit).
x!a
Recall that the existence of this limit does not depend on f (a).
2. The value of L must be equal to f (a), provided that f (a) exist:
i.e. a is in the domain of f (x).
Thus for f (x) to be not continuous at a, it could fail either 1 or 2.
Removable Discontinuity
In this scenario, condition 1 holds:
lim f (x) = L
x!a
exists
However, condition 2 does not hold:
f (a) 6= L
or
f (a) does not exist
In any case, we can redefine the function at a, by setting f (a) = L.
Then the function becomes continuous at point a.
We have ”removed” the discontinuity.
Removable discontinuity typically arises when an cancellation occurs in a
fraction (rational function).
For example:
(x 1)(x + 1)
f (x) =
x 6= 1
x 1
Note that f (1) does not exist. However, it is clear that lim f (x) = 2.
x!1
Thus we may redefine f (1) = 2. Then f (x) becomes:
(
(x 1)(x+1)
x 6= 1
x 1
f (x) =
2
x=1
is now continuous at a = 1. Note that f (x) = x + 1 everywhere.
Author: Xiao
Page 27
MAT137
7 TYPES OF DISCONTINUITY
Jump Discontinuity
In this scenario, condition 1 does not hold:
lim f (x) does not exist as 2 -sided limit
x!a
However, both 1-sided limits exist but are not equal:
lim f (x) = L1
lim f (x) = L2
x!a+
x!a
L1 6= L2
Note that these limits again does not depend on f (a).
Intuitively, f (x) ”jumps” from L1 to L2 as x goes acroos a.
Jump discontinuity typically arises when a function is piecewise defined.
For example:
(
1
x>0
|x|
=
f (x) =
x
1 x<0
We have lim f (x) =
x!0
1 and lim+ f (x) = 1, as a jump at x = 0.
x!0
Infinite Discontinuity
In this scenario, one or both of the 1-sided limits are infinite:
lim f (x) = ±1
x!a
or
lim f (x) = ±1
x!a+
or both
Infinite discontinuity typically arises when a fraction is dividing by 0.
However, we need to require that the numerator to be non-zero.
For example:
x 1
f (x) =
x 6= 1, 1
(x 1)(x + 1)
Note that
lim + f (x) =
x! 1
2
1
1
[+]
lim +
= lim +
=
=1
2 x! 1 x + 1 x! 1 x + 1
[+]
where a non-zero number dividing by 0 gives infinity.
lim+ f (x) =
x!1
1
x
lim+
2 x!1 x
1
1
1
= lim + · 1 =
1 x! 1 2
2
where 0 dividing by 0 does not gives infinity, and is called an indeterminant
form, which requires further manipulations. In this case, we ”cancel” (x 1)
on the numerator and denominator to attain the limit.
Author: Xiao
Page 28
MAT137
7 TYPES OF DISCONTINUITY
Essential Discontinuity
The above 3 scenarios covers most instances of discontinuity.
Any discontinuity that is not the above 3 scenarios is called an essential
discontinuity.
Essential discontinuity typically arises with limits similar to
1
lim sin
x!0
x
1
approaches infinity at a faster and faster pace.
x
The function sin(z) oscillate 1 period whenever the input increases by 2⇡.
1
The input of sin(z) in this case is z = , which increases faster and faster
x
toward infinity, passing through multiples of 2⇡ at an increasingly fast pace.
1
Thus sin oscillates faster and faster as x approach 0.
x
As x approach 0,
An ”intuitive” limit at x = 0 would be ”all numbers between 1 and 1”.
However, a limit must equal to 1 number only, not infinitely many numbers.
Thus
1
lim sin
does not exist
x!0
x
with no further classifications.
Similarly,
lim cos
x!0
Author: Xiao
1
x
does not exist
Page 29
MAT137
8
8 IDEA OF DERIVATIVE
Idea of Derivative
The derivative of f (x) at the point a is
f (a + h) f (a)
h!0
h
We take the limit of h approaching 0, so h 6= 0.
As h is on the denominator, we can not simply substitute h = 0 into the
expression. We need to perform algebraic manipulations to the fraction in
order to cancel the h on the denominator. Then we may substitue h = 0 to
attain the answer to f 0 (a).
f 0 (a) = lim
Note that f 0 (a) is typically di↵erent with di↵erent point a.
This can be understood as a function:
f (x + h) f (x)
f 0 (x) = lim
h!0
h
where x is now also thought of as the variable to the function f 0 .
There are several ways to intuitively understand the derivative f 0 (a):
1. Graph - Slope of f (x)
On the graph of y = f (x), m = f 0 (a) corresponds to the ⇣slope of
⌘ the
tangent line at the point a. The 2 points on the graph a, f (a) and
⇣
⌘
a + h, f (a + h) define a line, where x = a + h is either slightly left or
right of the point a. The slope of this line is the di↵erenece quotient
f (a + h) f (a)
f (a + h)
=
(a + h) a
h
f (a)
which appears in the definition of the derivative.
As we take the limit of h approaching 0, this line becomes a good
approximation of the graph near point a. Thus the slope of this
line approaches the slope of the f (x) at the point a.
The tangent line at x = a is defined to be the⇣line with
⌘ the slope as
0
the derivative f (a) passing through the point a, f (a) . Using pointslope form of the line:
y
Author: Xiao
f (a) = f 0 (a)(x
a)
Page 30
MAT137
8 IDEA OF DERIVATIVE
2. Rate of Change - Sensitivity to Input Changes
Consider an example f (x) = 2x where f is linear.
First set x = 1, we have f (1) = 2.
Now if we change the input by 1 unit, with x = 2, we have f (2) = 4.
We see that in this example, if we change the input by h units, the
output is changed by 2h units.
Typically if we change the input by a lot, we would expect the output to change a lot. However, the ratio of the changes between the
output and the input is the more meaningful information. This ratio
is the di↵erenece quotient
f (a + h) f (a)
= Average Rate of Change
h
The derivative measures this ratio when a very small change is done
to the input (h ! 0). We would expect a small change to the input
would result in a small change to the output. However, the ratio of the
changes between the output and the input may in fact be small or large.
A large ratio (positive or negative) indicates that the function f (x) is
very sensitive to changes to its input near the point a, where a small
change to the input will result in a (relatively) large change in output.
A ratio of 0 indicates that the function is not sensitive to changes to
its input near the point a, where a small change to the input will result
in an (extremely) small change in output.
By taking the limit of h approaching 0, while the change in input
becomes smaller and smaller, the ratio of the changes between the
output and the input may not become small.
In fact, the value which this ratio approaches to, is the derivative
f 0 (a), the instantaneous rate of change at point a, and it is the
’most accurate’ number to describe the sensitivity of input for f (x) at
the point x = a.
Because we take the limit of h approaching 0, the answer we attain
only meausres the sensitivity near the point x = a.
The derivative is a local phenomenon.
f 0 (a) does not give any information of f (x) at any other points.
Author: Xiao
Page 31
MAT137
8 IDEA OF DERIVATIVE
3. Velocity
Let’s use t in place of x for the variable.
Let f (t) = s(t) be the position of a particle at time t.
The average velocity between times t = a and t = b is
s(b) s(a)
di↵erence in position
=
b a
time used
This
correspond
⇣
⌘
⇣ to the
⌘ slope of the secant line defined by 2 points
a, s(a) and b, s(b) on the graph of s(t).
vavg =
The instantaneous velocity, or just simply velocity, at time t = a
is the derivative:
s(a + h) s(a)
s(t) s(a)
vinst = v(a) = s0 (a) = lim
= lim
t!a
h!a
h
t a
⇣
⌘
as the slope of the tangent line at a, s(a) .
The velocity can again be thought of as a function of time:
v(t) = s0 (t)
Question 12
Find the derivative f 0 (a) by definition.
1. f (x) = x3
1
2. f (x) = p
x
Question 13
Find the equation of the tangent line at the point given.
f (x) = x3 at (1, 1)
Question 14
1. Let f 0 (x) = 0 for all x and f (0) = 0. What can you say about f (2)?
2. Let f 0 (0) = 0 and f (0) = 0. What can you say about f (2)?
Question 15
Let s(t) = t2 6t.
Find the average velocity for t between [4, 8] and the instantaneous velocity
at t = 8.
Author: Xiao
Page 32
MAT137
9
9 THREE IMPORTANT THEOREMS
Three Important Theorems
Intermediate Value Theorem (IVT)
If f (x) is continuous on [a, b], and f (a) < 0, f (b) > 0,
then exists c 2 (a, b) with f (c) = 0.
Intuition:
To draw a continuous curve from below the x-axis to above the x-axis, the
curve must cross the x-axis.
Intermediate Value Theorem: Generalized Version
If f (x) is continuous on [a, b],
then for any K between f (a) and f (b), there exists c 2 (a, b) with f (c) = K.
Intuition:
To draw a continuous curve from y = f (a) to y = f (b), the curve must cross
all possible horizontal lines y = K between y = f (a) and y = f (b).
The theorem guarantees the existence of c, but gives no way of finding c.
Extreme Value Theorem (EVT)
If f (x) is continuous on [a, b],
then f (x) has maximum and minimum on the interval [a, b]:
1. There exists xM , with f (xM )
f (x) for all x 2 [a, b], as a maximum.
2. There exists xm , with f (xm )  f (x) for all x 2 [a, b], as a minimum.
Intuition:
On a closed interval [a, b], a continuous curve would have a highest point and
a lowest point.
Note that the assumption of closed interval is crucial:
If we only assume f (x) to be continuous on (a, b], then it is possible that
f (x) becomes infinite at a, in which case there would not exist a maximum.
The theorem guarantees f (x) would reach maximum and minimum, but the
maximum or minimum may be reached at more than 1 point. The theorem
again gives no way of finding the maximum and minimum.
Author: Xiao
Page 33
MAT137
9 THREE IMPORTANT THEOREMS
Mean Value Theorem (MVT)
If f (x) is continuous on [a, b], di↵erentiable on (a, b), then exists c 2 (a, b)
with
f (b) f (a)
f 0 (c) =
b a
Intuition:
Consider a car driving forward on a highway. Suppose the car travels 100km
forward in an hour. Can I conclude that at some time during the trip, the
velocity of the car is exactly 100km/h?
1. If the car always travels faster than 100km/h, then after 1 hour, it
would have travelled more than 100km.
2. Similarly, if the car always travels slower than 100km/h, then after 1
hour, it would have travelled less than 100km.
3. Thus there must be a time t1 where the velocity is below 100km/h,
and another time t2 where the velocity is above 100km/h. To change
the velocity from below 100km/h to above 100km/h, the car must pass
through the velocity of 100km/h at some time c between t1 , t2 .
Let f (x) = s(t) as the position of a car driving forward on a highway.
Then f 0 (x) = s0 (t) = v(t) is the velocity of the car.
Then Mean Value Theorem states (with the new notation):
There must exists c 2 (a, b) with
v(c) = s0 (c) =
s(b)
b
s(a)
Total distance
=
= average velocity
a
Total time
It is important to require f (x) to be di↵erentiable. In the above example,
the velocity of the car must change in a smooth and continuous manner.
The theorem guarantees the existence of c, but gives no way of finding c.
Question 16
Show that there are at least 3 solutions in
Estimate their values.
Question 17
Give the region of x with
(x
⇣ x ⌘3
⇡
2 sin x +
1
= 0?
2
x
> 0, by finding all zeroes and
1)(x + 1)
discontinuities of f .
Author: Xiao
Page 34
MAT137
10
10 DERIVATIVE
Derivative
f (a + h) f (a)
h!0
h
The function is di↵erentiable at point a if this limit exists as the derivative at a.
f 0 (a) = lim
f is di↵erentiable implies f is continuous.
Note that continuous does not imply di↵erentiable:
f (x) = |x| is continuous, but not di↵erentiable at x = 0.
Tangent Line
The point-slope form of line: y y0 = m(x x0 ) where (x0 , y0 ) is on the line.
The tangent line at (x0 , y0 ) = a, f (a) has slope m = f 0 (a).
Derivative Rules
dy
d
=
(y) (c)0 = 0 (f +g)0 = f 0 +g 0
(cf )0 = cf 0
dx
dx
f
f 0g g0f
(f g)0 = f 0 g + g 0 f
( )0 =
[f (g(x))]0 = f 0 (g(x)) · g 0 (x)
2
g
g
1
(xn )0 = n · xn 1
(ex )0 = ex
(ln x)0 =
x = eln(x) = ln(ex )
x
(sin(x))0 = cos(x) (cos(x))0 = sin(x)
sin(x)
1
1
cos(x)
tan(x) =
csc(x) =
sec(x) =
cot(x) =
cos(x)
sin(x)
cos(x)
sin(x)
y = f (x)
f 0 (x) =
(tan(x))0 = sec2 (x)
(sec(x))0 = sec(x)tan(x)
Inverse Function Theorem
(f
1 0
) (x) =
1
f 0 (f 1 (x))
⇡ ⇡
, ]
)
sin(✓) = x 2 [ 1, 1]
2 2
arccos(x) = ✓ 2 [0, ⇡]
)
cos(✓) = x 2 [ 1, 1]
1
1
arcsin0 (x) = p
arccos0 (x) = p
2
1 x
1 x2
⇡ ⇡
1
arctan(x) = ✓ 2 [
, ] ) tan(✓) = x 2 R arctan0 (x) =
2 2
1 + x2
arcsin(x) = ✓ 2 [
Author: Xiao
Page 35
MAT137
10 DERIVATIVE
Higher Order Derivatives
f 0 (x) is the first order derivative function of f (x), and we may take the
derivative of f 0 (x) a second time.
0
f 0 (x) = f 00 (x) = f (2) (x)
f 00 (x) is the second order derivative function of f (x).
Similarly, we may take even more derivatives on top of f 00 (x).
The n-th order derivative function of f (x),
f (n) (x)
is the result of taking the derivative n times onto the original function f (x).
Typically, there are no short-cuts or nice formulas for calculating n-th order
derivatives, one must proceed by taking one derivative at a time.
Position, Velocity, Acceleration
f (t) = s(t)
is the position of a particle at time t
s0 (t) = v(t)
is the velocity of a particle at time t
Then the second derivative of position
s00 (t) = v 0 (t) = a(t)
is the acceleration of a particle at time t
Question 18
Find the derivative.
1. f (x) =
x2 + 1
x2 1
3
2. f (x) = x tan(3x + 1)
r
x+2
3. f (x) =
x 1
Question 19
Find the n-th order derivative.
1. f (x) = ln(x)
Author: Xiao
4. f (x) = ex ln(ln(x))
5. f (x) = sin2 x + sin x2 + sin2 x2
6. f (x) = ex
e
7. f (x) = xx
2. f (x) =
p
x+1
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MAT137
10 DERIVATIVE
Question 20
1. Is there a continuous function on (0, 1) that takes on only 2 distinct
values?
2. Is there a continuous function on (0, 1) that does not have a maximum?
Question 21
Find conditions on A and B such that
(
x3
f (x) =
Ax + B
x1
x>1
f (x) is di↵erentiable at x = 1.
Question 22
Suppose
f (0) = 1, f 0 (0) = 2, f (1) = 0, f 0 (1) = 1, f (2) = 1, f 0 (2) = 1
g(0) = 2, g 0 (0) = 1, g(1) = 1, g 0 (1) = 0, g(2) = 1, g 0 (2) = 1
h(0) = 1, h0 (0) = 2, h(1) = 2, h0 (1) = 1, h(2) = 0, h0 (2) = 2
Find
1. (g f )0 (1)
2. (g f )0 (2)
3. (f
h g)0 (1)
4. (g f
h)0 (2)
Question 23
⇡
Find the equation of the tangent line of f (x) = cot(x) at a = .
6
Author: Xiao
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MAT137
10 DERIVATIVE
Exponential and Logarithms
We know the derivative of the natural exponential is (ex )0 = ex .
What if the base is not e but some other number a > 0?
We use the relation
a = eln(a)
⇣
⌘x
ax = eln(a) = ex·ln(a)
Thus any other exponential is just a horizontal stretch/shrink of ex by a
factor of ln(a). (Also a horizontal reflection if ln(a) < 0 with a < 1.)
By chain rule, and the above equality,
(ax )0 = ex·ln(a) · ln(a) = ax · ln(a)
1
We know the derivative of the natural logarithm is (ln x)0 = .
x
What if the base is not e but some other number a > 0?
We know the logarithm base a: loga (x) is the inverse function of ax .
y = loga (x)
,
x = ay
Using the relation a = eln(a) ,
y
⇣
x=a = e
ln(a)
⌘y
= ey·ln(a)
Taking ln on both sides, and using the fact that ln and e are inverse functions,
⇣
⌘
ln(x) = ln ey·ln(a) = y · ln(a)
Dividing by the constant ln(a),
y = loga (x) =
ln(x)
ln(a)
Thus any other logarithm is just a verticle stretch/shrink of ln(x) by a factor
1
of
. (Also a verticle reflection if ln(a) < 0 with a < 1.)
ln(a)
1
1
Using (ln x)0 = , with the extra constant of
in front,
x
ln(a)
⇣
⌘0
1
loga (x) =
x ln(a)
Author: Xiao
Page 38
MAT137
10 DERIVATIVE
Logarithmic Di↵erentiation
We know product rule as:
(f g)0 = f 0 g + f g 0
What if there are more than 2 functions multiplied together?
We could use the ’generalized product rule’ where:
(f gh)0 = f 0 gh + f g 0 h + f gh0
However, this rule becomes complicated if there are more than 3 functions.
There is an alternative method called logarithmic di↵erentiation, where
we use the facts:
x
ln(xy) = ln(x) + ln(y)
ln( ) = ln(x) ln(y)
ln(xy ) = y ln(x)
y
1. Given a product or quotient of functions, such as
y(x) =
f (x)2 g(x)
h(x)
Take ln of both sides to get
ln y(x) = ln
f (x)2 g(x)
h(x)
2. Apply the above rules, we get
ln y(x) = 2 ln f (x) + ln g(x)
ln h(x)
3. Take derivative against x on both sides using chain rule, we get
1
1
1
· y 0 (x) = 2
· f 0 (x) +
· g 0 (x)
y(x)
f (x)
g(x)
1
· h0 (x)
h(x)
4. Solve for y 0 (x) by multiplying y(x) to the right side,
h 1
i
1
1
y 0 (x) = y(x) 2
· f 0 (x) +
· g 0 (x)
· h0 (x)
f (x)
g(x)
h(x)
Question 24
Find the derivative y 0 (x) using logarithmic di↵erentiation.
p
(x2 + 1)3 (2x 5)2
x2 + 3
p
a) y(x) =
b)
y(x)
=
3
(x2 + 2)2
x2 + 2x
Author: Xiao
Page 39
MAT137
11
11 LINEAR APPROXIMATION AND DIFFERENTIAL
Linear Approximation and Di↵erential
⇣
⌘
Recall the tangent line is a good approximation of y = f (x) near a, f (a) :
y
f (a) = f 0 (a)(x
a)
Thus we may say, for x ”near” a, (meaning (x
a) should be small,)
y = f (x) ⇡ f (a) + f 0 (a)(x
a)
This is called the linear approximation of f (x) near a.
It is also called first order Taylor approximation of f (x) near a.
Di↵erential
We may rewrite the tangent line with a di↵erent notation.
The di↵erence in y =
y = f (x)
The di↵erence in x =
x=x
f (a)
a
( is the greek letter ”delta”, used to mean ”change in”.)
Then the above equation reads:
y ⇡ f 0 (a) ·
x
If we let the di↵erence to ”approach zero”, the approximation becomes better.
We say y becomes the di↵erential dy, x becomes the di↵erential dx:
dy = f 0 (a)dx
Now that if we ”switch notation” and write a instead of x, then
dy = f 0 (x)dx
dy
= f 0 (x)
dx
dy
Note that while both equations are similar in spirit,
is NOT a fraction,
dx
and can not be treated like one.
Question 25 p
Approximate 3.99.
Author: Xiao
Page 40
MAT137
12
12 INVERSE FUNCTION
Inverse Function
f :A!B
A is the domain of f , B is the codomain of f .
f is one to one if:
Any 2 di↵erent points in the domain will get sent (by f ) to 2 di↵erent points
in the codomain.
Equivalently, f is one to one if:
There can not exist 2 di↵erent points that get sent (by f ) to the same point.
The simplest logical statement for one to one is:
f is one to one
,
If f (a) = f (b)
then a = b
If it seems that there are 2 di↵erent points, a and b, which when put into f ,
result in the same value: f(a) = f(b) , it actually means the 2 points a and b
are the same point to begin with: a = b
Example
f (x) = x+3. Suppose f (a) = f (b), a + 3 = b + 3, so a = b ) f is one to one.
The image is all possible values that f can take.
image(f ) = {f (x) | x 2 A}
f is onto if: image = codomain.
Thus, we can easily make f onto, by simply defining B to be image(f ).
Example
Consider the (one to one) function,
f :R!R
f (x) = arctan(x)
Now at first, we can consider A = R (on the left), and B = R (on the right).
However, as x can be any real number in A = R, arctan(x) can take on
⇡ ⇡
every value in (
, ). (We can see this with the graph).
2 2
⇡ ⇡
Thus, image(f ) = (
, ).
2 2
Now if we set
⇡ ⇡
f :R!(
, )
2 2
Then f is one to one and onto.
Author: Xiao
Page 41
MAT137
12 INVERSE FUNCTION
Horizontal Line Test
Given a function f : f is one-to-one if and only if every possible horizontal line in the xy-plane intersect the curve at most once.
A function f is one to one and onto,
f :A!B
1
if and only if there exists an inverse function f
1
f
,
:B!A
The function f takes a 2 A as input, and gives b 2 B as output.
f (a) = b
The function f 1 takes the b 2 B as input (which is the output of f ), and
gives back a 2 A, as the output of f 1 .
f
1
Thus the domain of f
The inverse of f
1
1
(b) = a
is B, and the image of f
is f , written as (f
1
)
1
1
is A.
= f.
This gives a one to one correspondence between the points in A and B.
Any point in A is sent into B using f , and f 1 sends the result back into A.
f
1
f f
f (a) = a
1
(b) = b
for all a 2 A
for all b 2 B
In the previous example, f (x) = arctan(x), so f 1 (y) = tan(y), where
⇡ ⇡
f :R!(
, )
2 2
⇡ ⇡
f 1:(
, )!R
2 2
Although f 1 (y) is defined by a formula to be tan(y), it has a di↵erent
domain than the standard tan(y) function.
In particular, f (x) = arctan(x) and f 1 (y) = tan(y) are only inverses to
each other in the domain and image as defined above.
⇡ ⇡
For y 62 (
, ), we have arctan(tan(y)) 6= y.
2 2
Author: Xiao
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MAT137
12 INVERSE FUNCTION
Inverse Trignometric Functions
We know that sin(✓), cos(✓), tan(✓) are periodic with period 2⇡.
(We can even say tan(✓) has period ⇡.)
Consider f (✓) = sin(✓) for example.
Let
f :A=R!B=R
We have image(f ) = [ 1, 1], so we set
f : A = R ! B = [ 1, 1]
Then f is onto.
However, there are too many elements in the domain A of f , which get
sent to the same value in the codomain B of f .
For example, we have f (0) = 0 2 B, and f (2⇡) = 0 2 B, and f (4⇡) = 0 2 B.
This is called not one-to-one.
In order to make f one-to-one, we can restrict the domain, by removing
points in the domain that cause the above problem: f (0) = f (2⇡) = f (4⇡).
We (usually) choose an interval as the new domain A of f , to attain one-to-one.
f :A!B
The most common way for f (✓) = sin(✓) is to choose A = [
Then f is one to one and onto, and f
1
(x) = arcsin(x),
⇡ ⇡
, ].
2 2
⇡ ⇡
, ] ! [ 1, 1]
2 2
⇡ ⇡
f 1 (x) = arcsin(x) : [ 1, 1] ! [
, ]
2 2
f (✓) = sin(✓) : [
Similarly,
f (✓) = cos(✓) : [0, ⇡] ! [ 1, 1]
1
(x) = arccos(x) : [ 1, 1] ! [0, ⇡]
⇡ ⇡
f (✓) = tan(✓) : (
, )!R
2 2
⇡ ⇡
f 1 (x) = arctan(x) : R ! (
, )
2 2
f
Author: Xiao
Page 43
MAT137
12 INVERSE FUNCTION
Question 26
a) Find the following values.
2⇡
2⇡
2⇡
x = cos( ), y = sin( ), z = tan( ).
3
3
3
b) Verify whether the following quantities exists, and evaluate them.
arccos(x), arcsin(y), arctan(z).
c) Verify whether the following equations hold.
cos(arccos(x)) = x, sin(arcsin(y)) = y, tan(arctan(z)) = z.
d) Conclude that
cos(arccos(x)) = x
for all x in the domain of arccos, which is [ 1, 1].
Similarly,
sin(arcsin(x)) = x tan(arctan(x)) = x
for all x in the domain of arcsin (which is [ 1, 1]), and arctan (which is R).
e) Conclude that
arccos(cos(✓)) = ✓
for ✓ 2 [0, ⇡], the restricted domain of cos(✓).
Similarly,
arcsin(sin(✓)) = ✓
⇡ ⇡
for ✓ 2 [
, ], the restricted domain of sin(✓).
2 2
arctan(tan(✓)) = ✓
⇡ ⇡
for ✓ 2 (
, ), the restricted domain of tan(✓).
2 2
5⇡
⇡
7⇡
f) Evaluate arcsin(sin( )), arccos(cos(
)), arccos(cos( )).
6
3
6
Recall the sum of angles formula:
sin(x + y) = sin(x)cos(y) + cos(x)sin(y)
cos(x + y) = cos(x)cos(y)
p
sin(x)sin(y)
p
3
), arccos( 3),
1
g) Evaluate: arccos(1), arccos( ), arccos(
2
p2
p
1
3
arcsin(0), arcsin(1), arcsin( ), arcsin(
), arctan(1), arctan(
3)
2
2
Author: Xiao
Page 44
MAT137
12 INVERSE FUNCTION
Question 27
Sketch the graph of the following functions.
a) f (x) = sin(arcsin(x)), where x is a ratio.
b) f (x) = arcsin(sin(x)), where x = ✓ is an angle.
c) f (x) = tan(arctan(x)), where x is a ratio.
d) f (x) = arctan(tan(x)), where x = ✓ is an angle.
Question 28
Simplify the following expressions which are given as a composition of trignometric function and inverse trignometric function.
⇡ 3⇡
a) arccos(sin(✓)), where ✓ 2 [ , ] is an angle.
2 2
⇡ ⇡
Repeat with ✓ 2 [
, ].
2 2
b) sec(arccos(x)), where x is a positive ratio.
c) sin(arccos(x)), where x is a positive ratio.
d) sin(arctan(x)), where x is a positive ratio.
e) tan(arcsin(x)), where x is a positive ratio.
Author: Xiao
Page 45
MAT137
12 INVERSE FUNCTION
Solve for inverse function f
1
1. Set y = f (x)
2. Isolate x on one side: x = g(y), where g(y) is the right side with only y.
1
3. Write f
(y) = g(y), then switch all y to x, so f
1
(x) = g(x).
Note: The last step of ’switching y to x’ is optional.
The x after the switch is not the same x as we started out with.
This is used when we would like to write f 1 (x), with x being the variable.
Inverse Function Theorem
(f
1 0
) (x) =
1
f 0 (f 1 (x))
Question 29
f (x) =
(
x3
x2
1 x<0
x 0
a) Sketch the graph of f and verify using the graph that f is one to one.
b) Find f 1 , and then find (f 1 )0 (4) and (f 1 )0 ( 2).
Question 30
Find f 1 (x).
1. f (x) = x + 3
p
2. f (x) = 2
3x
3. f (x) =
2x + 1
x 1
4. f (x) =
5. f (x) =
1+e
1 e
ex
e
x
x
x
2
Question 31
1
2
Evaluate ln(e2x ), ln(2ex ), e2ln(x) , ln( 3 ), eln(ln(e )) .
e
Question 32
Solve for x.
1. e2
2. e2x
3x
=6
4. ln(x) + ln(x
1) = 1
3ex + 2 = 0
3. ln(ln(x)) = 1
Author: Xiao
5. 1 < e2x+1 < 2
Page 46
MAT137
13
13 IMPLICIT DIFFERENTIATION
Implicit Di↵erentiation
1. Replace all y with y(x)
2. Take derivative against x on both sides.
3. Isolate y 0 (x) =
dy
dx
Remember to use product rule and chain rule, when taking derivatives.
Example:
cos(xy) = 1 + sin(y)
1.
cos(x · y(x)) = 1 + sin(y(x))
2.
d
d
cos(x · y(x)) =
[1 + sin(y(x))]
dx
dx
3. Using chain rule:
d
dy
[x · y(x)] = 0 + cos(y(x)) ·
dx
dx
Using produict rule on the left side:
dy
dy
sin(x · y(x)) · [1 · y(x) + x · ] = cos(y(x)) ·
dx
dx
dy
Isolate
:
dx
dy
dy
sin(x · y(x)) · y(x) x · sin(x · y(x)) ·
cos(y(x)) ·
=0
dx
dx
dy
sin(x · y(x)) · y(x) + [ x · sin(x · y(x)) cos(y(x))] ·
=0
dx
dy
[ x · sin(x · y(x)) cos(y(x))] ·
= sin(x · y(x)) · y(x)
dx
dy
sin(x · y(x)) · y(x)
=
dx
x · sin(x · y(x)) cos(y(x))
sin(x · y(x)) ·
Question 33
dy
d2 y
Find
and 2 of
dx
dx
x2 + 4xy + y 3 + 5 = 0
at (x, y) = (2, 1).
Author: Xiao
Page 47
MAT137
14
14 RELATED RATES PROBLEM
Related Rates Problem
1. Assign symbols to all variables in the problem.
2. Write an equation that relate all the variables.
3. Identify which variables change with time.
If some quantity x changes as time passes, replace x with x(t).
4. Take derivative against t on both sides using implicit di↵erentiation.
5. Plug in the values given for the variables in the problem.
Example:
The volume of a cone is increasing at 30 m3 /min. The diameter and the
height are always equal. How fast is the height increasing when it is 10
meters?
1. v = volume, r = radius, h = height
2.
1
v = ⇡r2 · h
3
Using the assumption, diameter = 2r = h ! r =
h
2
1 h
1
v = ⇡( )2 · h = ⇡ · h3
3 2
12
3. As time passes, both volume and height will change.
v(t) =
4.
d
d 1
v(t) = [ ⇡ · (h(t))3 ]
dt
dt 12
1
d
dv
= ⇡ · (h(t))3
dt
12
dt
dv
1
dh
= ⇡ · 3 · (h(t))2 ·
dt
12
dt
5.
30 =
Author: Xiao
1
⇡ · (h(t))3
12
1
dh
dh
⇡ · 3 · (10)2 ·
= 25⇡ ·
12
dt
dt
)
dh
6
=
dt
5⇡
Page 48
MAT137
14 RELATED RATES PROBLEM
Question 34
The radius of sphere is increasing at the rate of 4 cm/s. How fast is the
volume increasing when the diameter is 80 cm?
Question 35
Two cars start at the same point. One travels west at 25 km/h, the other
travels south at 60 km/h. What is the rate of the distance between them
changing after 2 hours?
Question 36
A policeman is at 200m from a straight highway. Down the highway, 200m
from the point on the highway closest to the policeman, is an emergency call
box. The officer points the radar gun at the call box. A car passes the call
box, and the radar indicates the distance between the policeman and the car
is increasing at 72km/h. The speed limit is 90km/h. Is the car above the
speed limit?
Question 37
The water level in a hemispherical tank of radius 2 m is dropping at a rate
of 5 cm/sec when the depth of the water in the tank is 1 m. At what rate
is the radius of surface area of the water decreasing, when the depth of the
water in the tank is 1 m?
Question 38
A camera 3000 m from the launch pad is to be always aimed at a rocket as it
rises. How fast is the angle of elevation of the camera changing at the instant
the rocket is 4000 m above the launch pad, and rising at a rate of 880 m per
sec?
Question 39
A 2-m tall man walks along the side of an 8-m wide street, on which the
only lamppost is on the other side of the street. The man walks at a rate of
2 m/sec. The lamppost is 10 m tall. When he is 6 m from a point directly
opposite the lamppost, how fast is the length of his shadow changing?
Author: Xiao
Page 49
MAT137
15
15 MEAN VALUE THEOREM
Mean Value Theorem
Mean Value Theorem
If f is continuous on [a, b], di↵erentiable on (a, b),
then there exists c between (a, b) so that
f 0 (c) =
f (b)
b
f (a)
a
Corollary: Rolle’s Theorem
If furthermore, f (b) = f (a), then there exists c between (a, b) so that
f 0 (c) = 0
Increasing and Decreasing Function
f (x) is increasing on an interval [a, b], if for any x1 < x2 , we have f (x1 ) < f (x2 ).
f (x) is decreasing on an interval [a, b], if for any x1 < x2 , we have f (x1 ) > f (x2 ).
Results from Mean Value Theorem
If f 0 (x) = 0 for all x in an interval, then f (x) is a constant function.
i.e. There exists some constant c such that f (x) = c for all x in the interval.
(If f 0 (x) = g 0 (x), then f (x) g(x) = c is constant.)
If f 0 (x) 6= 0 for all x in an interval, then f (x) is one-to-one.
If f 0 (x) > 0 for all x in an interval, then f (x) is increasing.
If f 0 (x) < 0 for all x in an interval, then f (x) is decreasing.
Local Extrema
A point c is a local maximum if there exists an interval (c l, c + l) such
that for all x in the interval, f (x)  f (c) (when f (x) is defined).
A point c is a local minimum if there exists an interval (c l, c + l) such
that for all x in the interval, f (x) f (c) (when f (x) is defined).
If c is a local extrema in the interior of an interval [a, b],
then c is a critical point: f 0 (c) = 0 or f 0 (c) d.n.e.
First Derivative Test for Local Extrema
At a critical point c ( f 0 (c) = 0 or f 0 (c) d.n.e.) :
f 0 (x) < 0 on the left and f 0 (x) > 0 on the right ) c is a local minimum.
f 0 (x) > 0 on the left and f 0 (x) < 0 on the right ) c is a local maximum.
Author: Xiao
Page 50
MAT137
15 MEAN VALUE THEOREM
Question 40
Prove that
f (x) = arcsin
p
1 x
+ 2arctan x
1+x
is constant on some interval.
What is the constant?
What’s the largest interval where this is true?
Question 41
Find all functions defined on R that has f 0 (x) = 3.
Furthermore, show that no other function (other than those that you have
found) can satisfy f 0 (x) = 3.
Question 42
a) Suppose f 0 (x) has only 1 zero on [a, b].
How many zeroes can f have at most on [a, b]?
b) Suppose f 0 (x) has only 2 zero on [a, b].
How many zeroes can f have at most on [a, b]?
c) Suppose f 0 (x) has only k zero on [a, b].
How many zeroes can f have at most on [a, b]?
d) Suppose f 00 (x) has only 1 zero on [a, b].
How many zeroes can f have at most on [a, b]?
e) Suppose f (n) (x) has only 1 zero on [a, b].
How many zeroes can f have at most on [a, b]?
Question 43
Find the critical points, and classify them as local extremas.
1. f (x) = x4
2. f (x) = |x|
(
x2 + 1 x < 1
3. f (x) =
x+1 x 1
Author: Xiao
Page 51
MAT137
15 MEAN VALUE THEOREM
Question 44
Sketch the graph of
f (x) =
(
2x + 2 x 
x3 x x >
1
1
Find f 0 (x).
Determine whether f satisfies the conditions of the mean value theorem on
[ 3, 2].
Find the value c from the conclusion of the mean value theorem.
Question 45
Suppose f 0 (x)  M for all x, where M is a constant.
let a < b.
Show the rate of increase of f (x) is bounded by M :
f (b)  f (a) + M (b
Author: Xiao
a)
Page 52
MAT137
16
16 MAXIMUM AND MINIMUM
Maximum and Minimum
Extreme Value Theorem (EVT)
If f (x) is continuous on [a, b],
then f (x) has (absolute) maximum and minimum on the interval [a, b]:
1. There exists xM , with f (xM )
f (x) for all x 2 [a, b], as a maximum.
2. There exists xm , with f (xm )  f (x) for all x 2 [a, b], as a minimum.
To find the maximum and minimum of f (x) on [a, b] when f is di↵erentiable,
we can use the second derivative test:
1. Compute f 0 (x). Set f 0 (x) = 0.
The values of x that satisfy this equation are called critical points.
(Technically, if f 0 (x) doesn’t exist, then x is also a critical point.)
2. Compute f 00 (x). For each critical point a,
If f 00 (a) > 0, a is a local minimum.
If f 00 (a) < 0, a is a local maximum.
If f 00 (a) = 0, then it is unclear.
(We can also use the first derivative test for local max/min.)
3. Compare all values of f (a) for all critical points and endpoints of
interval. The largest value is absolute (global) maximum. The
smallest value is absolute (global) minimum.
We can carefully extend the above results to f (x) with domain being R.
1. For lim f (x) = +1 = lim f (x), then there exists absolute minimum.
x!1
2. For lim f (x) =
x!1
x! 1
1 = lim f (x), then there exists absolute maximum.
x! 1
3. For lim f (x) = L = lim f (x), where L 6= ±1:
x!1
x! 1
If there exists x1 with f (x1 ) > L, then there exists absolute maximum.
If there exists x2 with f (x2 ) < L, then there exists absolute minimum.
4. For lim f (x) = +1, lim f (x) =
x!1
x! 1
1 (infinities of opposite signs),
there are no absolute maximum and no absolute minimum.
Note: it is important to assume f (x) has domain being R, that is to say f (x)
can not be approaching infinity anywhere on R.
Author: Xiao
Page 53
MAT137
16 MAXIMUM AND MINIMUM
Example:
f (x) =
1.
x2 4
x2 + 4
2x · (x2 + 4) 2x · (x2 4)
2x3 + 8x 2x3 + 8x
=
(x2 + 4)2
(x2 + 4)2
16x
= 2
=0
(x + 4)2
The denominator is always positive, so the numerator is 0. The only
critical point is x = 0
f 0 (x) =
2.
f 00 (x) =
16 · (x2 + 4)2 16x · 2(x2 + 4) · 2x
(x2 + 4)4
16 · (x2 + 4) 64x2
64 48x2
=
(x2 + 4)3
(x2 + 4)3
At the critical point x = 0,
=
f 00 (0) =
64
>0
43
Thus x = 0 is a local minimum.
3. Compare the ”endpoints”, in this case:
lim f (x) = 1
x!±1
We do have f (0) = 1 < 1.
However, notice that f (x) < 1 since the numerator is always smaller.
So, there are no absolute maximum, but there exists absolute minimum.
Thus x = 0 is the absolute minimum.
Question 46
Find all critical points, local extrema, global extrama, interval of increase
and decrease.
a) f (x) = (x 1)2 (x 2)2
on the interval [0, 3]
b) f (x) = (
|x2 1|
on R
2
2 2x x
2x0
c) f (x) =
|x 2|
0<x3
Author: Xiao
Page 54
MAT137
17
17 OPTIMIZATION
Optimization
1. Identify the quantity that needs to be maximized/minimized. Set as y.
2. Identify the quantity that can be changed. Set as x.
3. Write an equation that relate all the variables.
Draw a diagram if needed.
4. Calculate
dy
, and set to 0.
dx
5. Solve for x, and the quantities needed.
Examples:
There are 1200cm2 of material to construct a rectangular container with
square base and with open top. What is the largest volume?
1. Set v = volume
2. Set b = base, h = height
3. v = b2 · h and 1200 = b2 + 4hb
Solving for h:
1200 b2
h=
4b
2
4
1200b
b
v=
= 300b
4b
4.
5.
dv
= 300
db
b3
4
3b2
=0
4
1200
= 400 ) b = 20
3
1200 400
h=
= 10
80
v = b2 · h = 400 · 10 = 4000
b2 =
Author: Xiao
Page 55
MAT137
17 OPTIMIZATION
Question 47
How much fencing is needed to define two adjacent rectangular playgrounds
of the same width and total area 15, 000m2 ?
Question 48
Find the point(s) on the parabola y = x2 closest to the point (3, 0).
Question 49
You want to sell a square open box: the bottom is a square, the four sides
are identical rectangles, and there isn’t anything on the top. The box should
have a volume of 1000 cm3 . The material for the bottom costs $2 per cm2
and the material for each of the four sides costs $3 per cm2 . You know you
will be able to sell each box for $1,250. Is this operation profitable?
Question 50
A farmer wants to hire workers to pick 1600 bags of beans. Each worker can
pick 10 bags per hour and is paid $1.00 per bag. The farmer must also pay a
supervisor $20 per hour while the picking is in progress. She has additional
miscellaneous expenses of $8 per worker (but not for the supersivor). How
many workers should she hire to minimize the total cost? What will the cost
per bag picked be?
Question 51
A power line is needed to connect a power station on the shore of a river
to an island 4 kilometers downstream and 1 kilometer o↵shore. Find the
minimum cost for such a line given that it costs $50,000 per kilometer to lay
wire under water and $30,000 per kilometer to lay wire under ground.
Question 52
A bus company runs a bus at a fare of $37 per person if 16 to 35 passengers
sign up for the trip. The company does not charter trips for fewer than 16
passengers. The bus has 48 seats. If more than 35 passengers sign up, then
the fare for every passenger is reduced by 50 cents for each passenger in excess of 35 that signs up. Determine the number of passengers that generates
the greatest revenue for the bus company.
Author: Xiao
Page 56
MAT137
18
18 CONCAVITY AND POINT OF INFLECTION
Concavity and Point of Inflection
When f 0 (x) > 0, f (x) is increasing at x.
When f 0 (x) < 0, f (x) is decreasing at x.
When f 00 (x) > 0, f (x) is concave up at x.
When f 00 (x) < 0, f (x) is concave down at x.
Concave up looks like a bowl right-side up sitting on the table.
Concave down looks like a bowl that’s put upside down.
A point x is a point of inflection if f 00 (x) = 0 and f 00 changes sign at x.
(If f 00 still changes sign at x, x is a point of inflection even if f 00 (x) d.n.e.)
Solve f 00 (x) = 0, and check whether f 00 (x) changes sign.
Example:
x2 4
x2 + 4
16x
f 0 (x) = 2
(x + 4)2
The denominator of f 0 (x) is always positive, so the sign of f 0 (x) depends
only on the numerator.
Thus f (x) is increasing for x > 0, and decreasing for x < 0.
f (x) =
f 00 (x) =
64 48x2
=0
(x2 + 4)3
The denominator of f 00 (x) is always positive, so set the numerator is 0.
64
4
2
=
!
x = ±p
64 48x2 = 0
!
x2 =
48
3
3
00
Note that f (x) has no verticle asymptotes, so f (x) always exists. We need
2
to check the 3 regions set out by x = ± p .
3
p
p
00
We have f (0) > 0, so f (x) is concave up between 2/ 3 and -2/ 3.
As x approaches +1 and 1, 64 48xp2 approaches p1.
Thus f (x) is concave down for x < 2/ 3 and x > 2/ 3
p
So x = 2/ 3 and x =
Author: Xiao
p
2/ 3 are points of inflection.
Page 57
MAT137
19
19 GRAPHING
Graphing
We can try to accurately sketch a graph by finding the following:
1. Domain
2. Verticle asymptotes: lim f (x) = ±1
x!a
(typically happens for x not in the domain)
3. Horizontal asymptotes: lim f (x) and lim f (x)
x!1
x! 1
Possible slant asymptote for rational functions:
am xm + ... + a0
f (x) =
bn xn + ... + b0
m=n+1
lim f (x) = ±1, but f (x) has a slant asymptote attained by long division.
x!±1
f (x) = kx + b +
r(x)
where kx + b is the slant asymptote.
q(x)
Find intersections to horizontal/slant asymptotes if possible.
4. Points of intersection with axis: Set x = 0, and y = 0
5. Take derivative: Intervals of increase/decrease, critical points f 0 (a) = 0.
6. Take 2nd derivative: Intervals of concavity, points of inflection.
Classify critical points as maximum, minimum, or neither.
7. Other properties: order of zeroes (for rational functions), discontinuity,
endpoint behaviour (for intervals), verticle tangent lines, corners...
Question 53
Sketch the graph.
1. f (x) = x2 (x + 2)3
2. f (x) =
x2
x2
1
3
3. f (x) =
Author: Xiao
x
+1
x2
4. f (x) =
ex
ex + 1
5. f (x) = x1/3
p
6. f (x) = x 1
7. f (x) = |2x
x2
x2 |
Page 58
MAT137
20
20 L’HOPITAL’S RULE
L’Hopital’s Rule
When
lim f (x)
0
=
lim g(x)
0
lim
OR
±
f (x)
f 0 (x)
= lim 0
g(x)
g (x)
1
1
Usually, we plug in the value of x and get 00 , so we take derivative and plug
in x again.
Example:
x3
0
lim 2
=
x!0 x + x
0
Taking derivative of the numerator and denominator,
3x2
0
= =0
x!0 2x + 1
1
= lim
Case of 0 · 1
lim[f (x) · g(x)] = 0 · 1
We can do one of two things:
lim[f (x) · g(x)] = lim
lim[f (x) · g(x)] = lim
f (x)
0
=
1
0
( g(x) )
g(x)
1
=
1
1
( f (x) )
Typically, the simpler fraction out of the two should be used to get the
answer.
Example:
1
sin(1/x)
0
lim x · sin( ) = lim
=
1
x!1
x!1
x
0
(x)
= lim
x!1
Author: Xiao
cos(1/x) ·
1
x2
1
x2
1
= lim cos( ) = cos(0) = 1
x!1
x
Page 59
MAT137
20 L’HOPITAL’S RULE
Case of 01 , 10 , 11 , ...
When the base of an exponent includes x, one must utilize the formula:
f (x)g(x) = (eln(f (x)) )g(x) = eg(x)·ln(f (x))
Example:
lim (tan(2x))x = lim+ ex·ln(tan(2x))
x!0+
x!0
Using property of limits, we can move the limit onto the exponent.
h
i
lim x · ln(tan(2x))
= e x!0+
So we only need to compute
lim+ x · ln(tan(2x)) = lim+
x!0
= lim+
ln(tan(2x))
x!0
1
tan(2x)
· sec2 (2x) · 2
1
x2
x!0
= lim+
x!0
1
x
cos(2x)
1
sin(2x) cos2 (2x)
1
x2
=
·2
ln(0)
1
=
1
1
= lim+
x!0
2x2
sin(2x)cos(2x)
The limit of cos(2x) is 1, so we can split up the limit to simplify calculation.
= lim+
x!0
2x2
1
2x2
0
· lim+
= lim+
·1=
sin(2x) x!0 cos(2x) x!0 sin(2x)
0
Using L’hopital:
4x
0
= =0
x!0 cos(2x) · 2
2
Thus the limit in the exponent is 0.
= lim+
lim ex·ln(tan(2x)) = elim x · ln(tan(2x)) = e0 = 1
x!0+
L’hopital rule should not be used when the limit can be solved by other
techniques, as derivative of function typically is more complicated than the
original function.
Question 54
An important limit - Compount interest (with 100% interest per year):
lim
n!1
Author: Xiao
⇣
1+
1 ⌘n
=e
n
Page 60
MAT137
20 L’HOPITAL’S RULE
Question 55
tan(3t)
t!0 ln(1 + 2t)
1. lim
h3
h!2 h3
5h2 + 3h + 6
h2 3h + 2
2. lim
1 cos(3t)
t!0 t · ln(1 + t)
3. lim
4. lim (x7
5x3 + 2)e
x!1
5. lim
(x
x!1
6. lim
p
x!0
x
1)sin(x)
ex cos(x)
x+1
x
1
3
7. lim x · tan( )
x!1
x
1
1
)
x!0 xsin(x)
xtan(x)
p
x4 + 5x3 x2
9. lim
x!1
x
8. lim (
sin(x)
x!1
x
10. lim
11. lim (cos(x))1/x
2
x!0
ln(2)
12. lim x 1 + ln(x)
x!1
13. lim+ x
p
x
x!0
14. lim (2
x!1
Author: Xiao
x)tan(⇡x/2)
Page 61
MAT137
21
21 IDEA OF INTEGRATION
Idea of Integration
Let’s first assume f (x) > 0.
We want the area under the curve of f (x) on [a, b] (and above the x-axis).
We split the area into many rectangles, and add up their total area.
Area of Rectangle = base · height P
P
The total area of the rectangles is
base · height =
bi · h i
where i runs over all rectangles in the partition P .
i
The way we pick the bases of the rectangles on [a, b] is called partition.
P : t0 < t1 < ... < tn
t0 = a
tn = b
Once we have a partition P , we can split the rectangles into even smaller
rectangles (by adding some additional points in the list of ti ). This is called
a refinement of P .
But, how do we choose the height of the rectangles?
There are many methods to choose the height of the rectangles.
The following 2 methods for choosing height are important for theoretical purposes:
For Upper Sum, pick the height of a rectangle to be the maximum value
(actually the supremum) of f in that rectangle.
The total area of the rectangles computed with this method is called Uf (P ).
For Lower Sum, pick the minimum value (the infimum) of f instead.
The total area of the rectangles computed with this method is called Lf (P ).
However, finding maximum (supremum) and minimum (infimum) of the functions in many small rectangles are typically very difficult.
The following 2 methods are easy for computation:
For Right-point method, pick the height of a rectangle to be the function
value of the right most endpoint of the rectangle.
For Left-point method, pick the height of a rectangle to be the function
value of the left most endpoint of the rectangle.
Author: Xiao
Page 62
MAT137
21 IDEA OF INTEGRATION
Given a partition of [a, b], where t0 = a and tn = b
P : t0 < t1 < ... < tn
We may compute the total area explicitly.
In the i-th rectangle with base [ti 1 , ti ],
the right most endpoint is ti and the left most endpoint is ti 1 .
The total area computed by the Right-point method is:
Total area =
n
X
i=1
bi · h i =
n
X
(ti
i=1
ti 1 ) · f (ti )
The total area computed by the Left-point method is:
Total area =
n
X
i=1
bi · h i =
n
X
(ti
i=1
ti 1 ) · f (ti 1 )
Notice that the formula for the Right-point method is slightly easier.
If we furthermore assume that we split partition the interval [a, b] into
n equal length intervals, then we attain a formula for ti :
b a
Each rectangle would have base length bi =
.
n
b a
Since t0 = a, we must have t1 = a + bi = a +
.
n
b a
Then we must have t2 = t1 + bi = a + 2
.
n
Thus the pattern continues and we attain:
ti = a + i
b
a
n
The formula for the total area with Right-point method becomes:
n
n
X
X
b a ⇣
b a⌘
Total area =
bi · h i =
·f a+i
n
n
i=1
i=1
The formula for the total area with Left-point method becomes:
n
n
X
X
b a ⇣
b a⌘
Total area =
bi · h i =
· f a + (i 1)
n
n
i=1
i=1
Author: Xiao
Page 63
MAT137
21 IDEA OF INTEGRATION
Intuitively, as we use more rectangles, the total area of the rectangles would
become a better approximation to the true area under the curve of f (x).
Thus an informal definition of the integral is defined to be the true area
under the curve of f (x), approximated using more and more rectangles:
True Area =
Z
b
f (x)dx = lim
a
n!1
n
X
i=1
bi · h i
where hi corresponds to some (potentially unknown/undeterministic) way of
choosing the height of the rectangles. However, the only requirement is that
hi needs to be the function value of a point xi in the i-th rectangle. That is
to say hi must equal to f (xi ) for some xi 2 [ti 1 , ti ].
These types of sum (where we don’t specify the exact way of choosing base
and height) is called a Riemann Sum.
We may intuitively think that as n becomes large, bi becomes very narrow,
so it is centered around a point x. Then we can change notation bi ! dx.
If dx is narrow enough, the function values in the small rectangle are almost
identical, so we may set hi ! f (x).
Then informally: The ’sum’ becomes the ’integral’ sign.
This definition extends to the case where f (x) may also be negative.
In this case, we may still use the Riemann Sum formula, but we would be
approximating the ’Net area’:
Net area =
Z
b
f (x)dx = lim
a
n!1
n
X
i=1
bi · h i
where
1. When f (x) > 0, we count the area above the x-axis and below f (x) to
be ’positive area’.
2. When f (x) < 0, we count the area below the x-axis and above f (x) to
be ’negative area’.
The ’Net area’ is the sum of both the ’positive’ and ’negative’ areas.
Author: Xiao
Page 64
MAT137
21 IDEA OF INTEGRATION
When computing the total area of the rectangles using Riemman Sums when
f (x) are simple polynomials, the following formulas are helpful:
n
X
i=1
n
X
n(n + 1)
i=
2
n
X
i=1
c · s(i) = c ·
i=1
n
X
i=1
n(n + 1)(2n + 1)
i =
6
s(i)
2
n
X
i=1
⇥
n
X
i3 =
i=1
⇤
s(i) + t(i) =
n
X
i=1
n2 (n + 1)2
4
s(i) +
n
X
t(i)
i=1
where s(i), t(i) are expressions involving i, and c is a constant not involving i.
Question 56
Compute the following integral by using the right-point method with the
partition being n equal length intervals.
Z 1
Z 2
1.
2x dx
2.
2x2 x dx
0
1
Question 57
Let f (x) = x2 2 and x 2 [0, 3] be partitioned into 3 equal length intervals.
Compute the Riemann Sum explicitly, with the midpoint method, where
we choose the height to be the function value of the midpoint in each interval.
Distance and Displacement
The velocity v can be positive or negative.
The speed |v| is the absolute value of velocity and is always positive.
If speed is constant, then total distance travelled = speed · time.
If speed is not constant, we can split the time interval [a, b] into many small
time intervals, each with speed |v(t)| changing with time. We use the integral
to add up the distance travelled in each small time interval to attain the total
distance travelled,
Z b
Total distance travelled with t 2 [a, b] =
|v(t)|dt
a
If we integrate v, then we are also considering the interval when v is negative.
When v > 0, we are moving forward. When v < 0, we are moving backward.
The displacement is the change in position,
Z b
Total displacement with t 2 [a, b] =
v(t)dt
a
Author: Xiao
Page 65
MAT137
22
22 1D INTEGRATION
1D Integration
We want the area under the curve of f (x) on [a, b].
We split the area into many rectangles, and add up their total area.
The way we pick the base of the rectangles on [a, b] is called partition.
Once we have a partition P , we can split the rectangles into more smaller
rectangles. This is called a refinement of P .
For theoretical purposes, the following 2 methods of choosing height are important:
For Upper Sum, pick the height of a rectangle to be the maximum value
(actually the supremum) of f in each rectangle.
The total area of the rectangles computed with this method is called Uf (P ).
For Lower Sum, pick the minimum value (the infimum) of f instead.
The total area of the rectangles computed with this method is called Lf (P ).
Area of Rectangle = base · height P
P
The total area of the rectangles is
base · height =
bi · h i
i
where the index i runs over all rectangles in the partition P.
The total area of the rectangles using Upper Sum will always be larger than
(or equal to) the true area. Similarly, the total area using Lower Sum will
always be smaller (or equal) to the true area.
Thus, the true area is always sandwiched between Upper Sum and Lower
Sum. In fact, even if 2 di↵erent partitions are used, this is still true:
Result 1: For any partitions P1 and P2 ,
Lf (P1 )  ”T rue Area”  Uf (P2 )
By picking more rectangles, we can make the upper sum and lower sum closer
and closer together:
Result 2: If P2 is a refinement of P1 (P2 has all the rectangles of P1
and some more), then
Lf (P1 )  Lf (P2 )  ”T rue Area”  Uf (P2 )  Uf (P1 )
Author: Xiao
Page 66
MAT137
22 1D INTEGRATION
Definition 1: A function is integrable if the Upper Sum and Lower Sum
can made as close as you want (by picking more rectangles):
8✏ > 0, 9 P such that Uf (P ) Lf (P ) < ✏
Notice that the same partition P is used here in upper and lower sum.
What is the ”True Area”?
If we really can squeeze the upper sum and lower sum together, then there
is one intuitive answer.
The ”True Area” must be the ”minimum” of all possible upper sum, where
we try every possible way of picking the bases, and calculate the area using
the maximum (supremum) value.
It should also be equal to the ”maximum” of all possible lower sum.
To get around the fact that minimum or maximum might not exist:
Definition 2: R
b
”True Area” = a f = infimum of all possible {Uf (P )} = I ba
= supremum of all possible {Lf (P )} = I ba
Supremum and Infimum
A number u is the supremum of a set S if it is the least upper bound of S.
i.e. u is the smallest number which can satisfy:
For all s 2 S, s  u. (u is larger than everything in S)
A number l is the infimum of a set S if it is the greatest lower bound of S.
i.e. l is the largest number which can satisfy:
For all s 2 S, l  s. (l is smaller than everything in S)
The supremum is intuitively understood as the maximum of the set S.
The infimum is intuitively understood as the minimum of the set S.
Result 3: ✏ characterization of supremum and infimum
For all ✏ > 0, there exists s 2 S such that u ✏ < s  u.
(As u is the ’maximum’ of S, moving down even a tiny amount will bump
into an element of S.)
For all ✏ > 0, there exists s 2 S such that l  s < l + ✏.
(As l is the ’minimum’ of S, moving up even a tiny amount will bump into
an element of S.)
Author: Xiao
Page 67
MAT137
22 1D INTEGRATION
Question 58
Show whether the following functions f are integrable.
1. f (x) = 2 for x 2 [0, 2].
(
2 x 6= 1
2. f (x) =
for x 2 [0, 2].
4 x=1
(
2 x<1
3. f (x) =
for x 2 [0, 2].
4 x 1
8
>
<2 x < 1
4. f (x) = 4 1  x < 2 for x 2 [0, 2].
>
:
6 x=2
5. f (x) = x for x 2 [0, 2].
(
x
x<1
6. f (x) =
for x 2 [0, 2].
x+2 x 1
7. Suppose(g(x) is integrable and bounded (by M) on [a, b].
g(x) x 6= c
f (x) =
where f (c) 6= g(c).
f (c) x = c
8. Suppose g(x) is integrable and bounded (by M) on [a, b].
Let f (x) = g(x) except at finitely many points {c1 , ..., cn } where
f (ci ) 6= g(ci ).
(
1 x2Q
9. f (x) =
for x 2 [0, 1].
0 x 62 Q
10. Bonus:8
<1
f (x) = q
:0
Author: Xiao
p
written as an irreducible fraction
q
for x 2 [0, 1].
x 62 Q, x = 0, x = 1
x=
Page 68
MAT137
22 1D INTEGRATION
Question 59
Suppose f is increasing on [a, b]. Prove f is integrable.
(Note that we do not assume f is continuous.)
Now, suppose f is decreasing on [a, b]. Prove f is integrable.
A function f which is only increasing or only decreasing is called monotonic.
Question
60 R
Rb
b
Let a f (x)dx < a g(x)dx.
Let P be a partition of [a, b].
Are the following statements true or false? Justify your answers.
(Prove the statement is true, or give a counterexample if it is false.)
1. Lf (P ) < Ug (P )
2. Lf (P ) < Lg (P )
Rb
3. Lf (P ) < a f (x)dx
4. Uf (P ) < Ug (P )
Rb
5. a f (x)dx < Ug (P )
Rb
6. Uf (P ) < a g(x)dx
Question 61
Prove
inf f (x) · (b
x2[a,b]
a) 
Z
b
a
f (x)  sup f (x) · (b
a)
x2[a,b]
Question 62
Suppose f is integrable, Prove |f | is integrable.
Is it true that: If |f | is integrable, then f is integrable?
Question 63
Rb
Suppose f is continuous, f 0, a f = 0.
Prove f (x) = 0 for all x 2 [a, b].
Is this still true if we do not assume f is continuous?
Author: Xiao
Page 69
MAT137
23
23 FUNDAMENTAL THEOREM OF CALCULUS
Fundamental Theorem of Calculus
Fundamental Theorem of Calculus I
Let f (x) be continuous, constant a. Consider the function:
Z x
F (x) =
f (t) dt
a
(Notice the integrand is the function f , except with variable changed to t.)
Then
F 0 (x) = f (x)
(The derivative is the integrand, replace all values of t with x).
Combining with Chain Rule:
Z g(x)
F (x) =
f (t) dt
F 0 (x) = f (g(x)) · g 0 (x)
a
(Replace all t with g(x), then multiply g 0 (x).)
Fundamental Theorem of Calculus II
For f continuous, let F 0 (x) = f (x):
Z b
f (x) = F (b) F (a)
a
To compute an integral, we only need to find an anti-derivative.
i.e. Find a function F (x) such that F 0 (x) = f (x).
So we ’define’ the indefinite-integral to be one of the anti-derivative.
Z
f (x) = F (x)
There are infinitely many anti-derivative of f (x), they are di↵erent up to an
additive constant.
Common anti-derivatives
Z
1
xn =
xn+1
n+1
Author: Xiao
Z
x
e =e
x
Z
1
= ln |x|
x
Page 70
MAT137
23 FUNDAMENTAL THEOREM OF CALCULUS
Properties of Integrals
For f is integrable:
Z b
Z b
Z b
(f + g) =
f+
g
a
Z
b
f=
a
Z
a
a
Z
a
f
b
Z
b
f 
a
Z
Z
c
f=
a
b
b
cf = c
a
Z
b
f+
a
Z
Z
b
f
a
c
f
b
|f |
a
Since f (x) is integrable, f (x) must be bounded:
There exists m and M such that m  f (x)  M for all x 2 [a, b].
(Intuitively, bounded means f (x) is not approaching infinity.)
Z b
m(b a) 
f  M (b a)
a
If f  g on all of [a, b], then
If f is odd: f ( x) =
f (x),
If f is even: f ( x) = f (x),
Z
b
f
a
Z
Z
Z
g
a
a
f =0
a
a
f =2
a
b
Z
a
f
0
Mean Value Theorem for Integrals
For f (x) continuous, there exists c 2 [a, b] such that
Z b
f (x) = f (c)(b a)
a
Some Conditions that guarantee f is integrable on [a, b]
1. bounded and monotone (only increasing or decreasing on [a, b])
2. bounded and continuous
3. bounded and discontinuous only at finitely many points
If f (x) is approaching infinity, we are not able to integrate f (x) (at this time).
Author: Xiao
Page 71
MAT137
23 FUNDAMENTAL THEOREM OF CALCULUS
Function Definition using Integral
p
x2
arccos x = x 1
ln x =
Question 64
Find the derivative of F (x).
1.
Z
F (x) =
2.
F (x) =
3.
F (x) =
4.
F (x) =
Question 65
Consider
Z
Z
Z
+2
x
1
x
Z
p
1
t2 dt
x
1
dt
t
x2 +1
a
4
ex
2
dt
1 + et
5
dt
1 + cos(t)
x3 +x2
x2 x
1
1
dt
1 + sin2 (t)
5
Z
Z
1
dt
1 + cos(t3 )
2
x+1
0
Find c 2 [0, 2] in the conclusion of the mean value theorem for integral.
Question 66
Let g be an unknown di↵erentiable function.
Define
Z x
G(x) =
g(t)dt
0
Find
lim
x!0
Author: Xiao
R G(x)
0
arctan(s + 2s2 )ds
x2
Page 72
MAT137
23 FUNDAMENTAL THEOREM OF CALCULUS
Question 67
Evaluate the integral with FTC II (by finding anti-derivative).
Z 1
1.
2x 3
0
2.
3.
4.
5.
6.
7.
8.
9.
Z
Z
Z
8
Z
Z
Z
x
0
5
1
⌘
d ⇣p
1 + x2
dx
4
f (x) where f (x) =
0
Z
Z
p
3
2
2
1
x2
3
1
x
1
1
3
3
3
2x 0  x  1
x 1<x4
1
x2
3
1
(
1
x
1
x
Question 68
Find f (x) where f 0 (x) = x2 and f (2) = 1.
Question 69
Find the area bounded by curves x + y 2
Author: Xiao
4 = 0 and x + y = 2.
Page 73
MAT137
24
24 INTEGRATION TECHNIQUES
Integration Techniques
We would like to compute the integral (anti-derivative) of a function
Z
f (x)dx
Substitution (for indefinite integral: finding anti-derivative)
1. Set a portion of f (x) to be equal to u = u(x)
2. Compute
du
= u0 (x)
dx
3. Rearrange for dx regarding the left side as a fraction
dx =
du
u0 (x)
4. Replace dx in integral with the du expression above.
Replace the portion in f (x) you set to be u.
Hopefully, there is no more quantities involving x.
5. If there still are quantities involving x, solve x in terms of u
x = x(u)
Replace all remaining x in integral with x(u).
6. Once all quantities in integral are in terms of u, carry out the new
integral with u being the integration variable.
7. Substitute x back into u, to get the final answer in terms of x.
When to use substitution
Given f (x), try to set the most troublesome piece of f (x) to be u. Hopefully,
u0 (x) will be on the outside of the function f (x), directly multiplied to dx.
In this case, the quantity u0 (x) will cancel in the substitution process.
Note: When the bounds of the integral is given (definite integral), one can
also compute the new bounds for u, from the bounds given in x. However,
it is not necessary to compute the new bounds for u. Simply find the antiderivative in x, and plug in the bounds of x at the end.
Author: Xiao
Page 74
MAT137
24 INTEGRATION TECHNIQUES
Integration by Parts
1. Given f (x), attempt to split the function into 2 pieces, u and dv
Z
Z
f (x) = u · dv
Note: After choosing the quantity u, dv must be the rest of the function f (x)
2. Compute the derivative and the anti-derivative of u and dv
Z
du
0
0
= u (x) ) du = u (x) · dx
v = dv
dx
(dx here can be ommited.)
3. Apply formula
Z
Z
u · dv = u · v
du · v
4. Carry out the new integral on the right
When to use integration by parts
When f (x) is of the form xn sin(x), xn cos(x), xn ex , set u = xn , then du would
have one power less than before. Continue the same process until there is no
more powers of x left.
When f (x) contains functions with a lot of letters such as ln(x), arctan(x),
arccos(x), arcsin(x), set u to be this quantity. Then du would be only a
fraction, with no letters involved.
Example
u=x
Z
du = 1 · dx
x · cos(x) = u · v
Author: Xiao
Z
x · cos(x)
v=
Z
dv = cos(x)
cos(x) = sin(x)
Z
Z
du · v = x · sin(x)
1 · sin(x) dx
= x · sin(x) + cos(x)
Page 75
MAT137
24 INTEGRATION TECHNIQUES
Question 70
Z
1.
e 5x+1 dx
2.
3.
4.
5.
6.
7.
Z
Z
Z
Z
Z
Z
x
2
p
⇡/4
+ 1 dx
sin x
dx
cos3 x
0
2
x3
p
x
1(x + 1) dx
2.
3.
4.
5.
6.
7.
Z
Z
Z
Z
Z
9.
10.
11.
1
p
x2 2 + x dx
1/2
1/4
1
dx
x ln x
sin x cos x
dx
sin x + cos x
Question 71
Z
1.
xex dx
Z
8.
x sin(2x) dx
x2 cos x dx
12.
13.
Z
Z
Z
Z
14.
8.
Z
9.
10.
11.
ln x dx
12.
2
x ln x dx
Author: Xiao
Z
Z
ex cos x dx
sin(ln x) dx
Z
13.
Z
Z
Z
Z
Z
p
cos x
p
dx
x
p
x
x2
4
x2
dx
2x 7
dx
7x + 10
x3
dx
(x + 1)2
x2
x3
2x
3x2
5
dx
x+2
dx
x2 + 1
2x2
x
dx
x+2
arcsin x dx
x arctan x dx
arctan
p
x dx
x sec2 x dx
sec x dx
sec3 x dx
Page 76
MAT137
24 INTEGRATION TECHNIQUES
Question 72
Let v(t) = 2t 2. Find the total displacement and total distance travelled,
in the time interval [0, 3].
Question 73
Evaluate the limit by viewing it as a right endpoint Riemann Sum on [0, 1].
n i2
P
1. lim
n!1 i=1 n3
2. lim
n
P
n!1 i=1
i2
i
+ n2
Question 74
Find the area under the curve.
p
1. y = x x2 + 1 on 0  x  1
2. y = 2 sin x + 2 on 0  x  ⇡
Author: Xiao
Page 77
MAT137
25
25 AREA
Area
Given y = f (x) > 0 on x in [a, b], the area under the curve y = f (x) (and
above the x axis) is given by
Z b
Z b
Area =
f (x) dx =
y dx
a
a
If y = f (x) > g(x), then the area between the 2 curves is given by the
larger area minus the smaller area.
Z b
Z b
Area Di↵erence =
f (x) dx
g(x) dx
a
a
This can be extended to the case when the functions are negative as well.
Integration with Respect to y
Sometimes, we are given a curve implicitly, such as y 5 + y + 1 = x. It may
be difficult to solve for y = f (x), and sometimes we can not solve y = f (x).
In these scenarios, we can still compute area by taking integral against y.
Given x = f (y) > 0 on y in [c, d], the area between the curve x = f (y)
and the y-axis is given by
Z d
Z d
Area =
f (y) dy =
x dy
c
c
If x = f (y) > g(y) which means the curve f (y) is to the right of g(y) ,
then the area between the 2 curves is given by the larger area minus the
smaller area.
Z d
Z d
Area Di↵erence =
f (y) dy
g(y) dy
c
c
This can be extended to the case when the functions are negative as well.
Question 75
1. Find the total area between y = x3
2. Find the area between y = x
3. Find the area between y 2 + 3y
Author: Xiao
2x + 2 and y = x2 + 2.
1 and y 2 = 2x + 6.
x = 16 and |3y|
x = 0.
Page 78
MAT137
26
26 VOLUME
Volume
Volume by revolution
There are 2 quantities, radius r and height h. Depending on the rotation
axis, they will correspond to x and y.
The relation between x and y transform into relation between r and h.
Rotation axis as y-axis
r=x
h=y
Given x = f (y)
)
r = f (y).
We can compute the volume bounded between x = f (y) and the y-axis, by
adding up many horizontal circles on di↵erent values of y in (c, d).
Z
Z d
2
V = ⇡r dy =
⇡[f (y)]2 dy
c
Given y = f (x)
)
h = f (x).
We can compute the volume bounded between y = f (x) and the x-axis, by
adding up many thin cylinders on di↵erent values of x in (a, b).
Z
Z b
V = 2⇡r · h dx =
2⇡x · f (x) dx
a
Rotation axis as x-axis
h=x
r=y
Given y = f (x)
)
r = f (x).
We can compute the volume bounded between y = f (x) and the x-axis, by
adding up many verticle circles on di↵erent values of x in (a, b).
Z
Z b
2
V = ⇡r dx =
⇡[f (x)]2 dx
a
Given x = f (y)
)
h = f (y).
We can compute the volume bounded between x = f (y) and the y-axis, by
adding up many thin cylinders on di↵erent values of y in (c, d).
Z
Z d
V = 2⇡r · h dy =
2⇡y · f (y) dy
c
In both cases, we can take the di↵erence in volume between 2 curves.
Author: Xiao
Page 79
MAT137
26 VOLUME
Question 76
1. Consider the region bounded by x = y 3 , x = 8, y = 0.
Find the volume after revolving this region around y-axis.
2. Consider the region bounded by y = x2 , y = x1/3 .
Find the volume after revolving this region around y-axis.
p
3. Consider the region bounded by y = x, x + y = 6, y = 1.
Find the volume after revolving this region around x-axis.
4. Consider the region bounded by y = x2 , y = 9.
Find the volume after revolving this region around x-axis.
5. Find the volume of a cylinder, and a cone of the same dimensions.
6. Find the volume of a sphere.
7. Consider the region bounded by y = x2 + 4, y = 6x x2 .
Find the volume after revolving this region around x-axis, and y-axis.
p
p
8. Consider the region bounded by x = 0, y = 0, x + y = 1.
Find the volume after revolving this region around x = 3.
Do this in 2 ways: Integrate with respect to x, and with respect to y.
9. Consider a hemispherical bowl of radius R. We place a small iron ball
with radius r inside the bowl, where 2r  R (so it is completely inside
the bowl). Let the water level be h. Find the total volume of water.
Note, there are 2 cases:
If the ball is completely submerged, we subtract the ball’s total volume.
If the ball is only partially submerged, then we need to subtract that
portion of its volume.
10. Find the volume of a square based pyramid.
(Add area of squares instead of circles.)
11. Two cylinders with radius r intersect with their axis at right angle.
Find the volume of the intersection (that is inside both cylinders).
Strategy: Take the 2 cylinders to be x2 + z 2 = r2 and y 2 + z 2 = r2 .
Each horizontal section at a constant value of z is a square. Add up
all areas of the square at all values of z to attain the volume.
Author: Xiao
Page 80
MAT137
27
27 TRIGNOMETRIC INTEGRALS
Trignometric Integrals
Multiples of sin(x) and cos(x)
Let’s start with the basics:
Z
sin(x) = cos(x)
Z
cos(x) = sin(x)
Only 1 power of sin(x) or cos(x)
Use subtitution, set u to be the one with more powers.
Z
sin(x) · cos100 (x)dx
Z
u = cos(x)
sin(x) · cos100 (x)dx =
Z
du =
sin(x)dx
u101
=
101
u100 du =
cos101 (x)
101
One of sin(x) or cos(x) have odd power
Use the formula sin2 (x) + cos2 (x) = 1 to interchange cos2 (x) and sin2 (x) of
the odd power so that one power remain.
Z
Z
5
100
cos (x)sin (x) = cos(x)cos4 (x)sin100 (x)
=
=
=
Z
Z
Z
cos(x) 1
cos(x) 1
cos(x)sin100 (x)
sin2 (x)
2
sin100 (x)
2sin2 (x) + sin4 (x) sin100 (x)
Z
Z
102
2 cos(x)sin (x) + cos(x)sin104 (x)
Solve each integral with the method of 1 power of sin(x) or cos(x).
Note: This method also includes when one of sin(x) or cos(x) is missing.
Z
Z
Z
2
5
4
sin (x) = sin(x)sin (x) = sin(x) 1 cos2 (x)
=
Author: Xiao
Z
sin(x)
2
Z
2
sin(x)cos (x) +
Z
sin(x)cos4 (x)
Page 81
MAT137
27 TRIGNOMETRIC INTEGRALS
Only sin2 (x) or cos2 (x)
Use the sum of angles formula:
cos(2x) = cos(x + x) = cos(x)cos(x)
sin(x)sin(x) = cos2 (x)
sin2 (x)
Interchange the cos2 (x) and sin2 (x) as before, keep the one you need.
sin2 (x)
sin2 (x)
cos(2x) = cos2 (x)
1
cos(2x) = 1
2sin2 (x)
cos(2x) = 2cos2 (x)
1
cos(2x) = 1
1
sin2 (x) =
Z
Z
1
2
sin (x) =
cos(2x)
2
cos(2x)
2
Both of these integral are easy.
cos2 (x)
1 + cos(2x)
cos2 (x) =
2
Z
Z
1 + cos(2x)
cos2 (x) =
2
Only even powers of sin(x) and cos(x)
Use the above formula to interchange all even powers into powers of cos(2x)
Z
2
4
sin (x)cos (x) =
1
=
8
Z
(1
Z ✓
1
cos(2x)
2
◆✓
1 + cos(2x)
2
◆2
cos(2x)) 1 + 2cos(2x) + cos2 (2x))
Z
1
=
1 + 2cos(2x) + cos2 (2x) cos(2x) 2cos2 (2x) cos3 (2x)
8
Collect like terms, and replace the even powers again with the above formula
note that the term is now cos2 (2x). Integrate the odd powers as before.
Z
1
=
1 + cos(2x) cos2 (2x) cos3 (2x)
8
✓
◆
Z
1
1 + cos(4x)
=
1 + cos(2x)
cos(2x) 1 sin2 (2x)
8
2
For the last integral, use the substitution u = sin(2x) as before.
Author: Xiao
Page 82
MAT137
27 TRIGNOMETRIC INTEGRALS
Multiples of sec(x) and tan(x)
Let’s start with the basics:
d
sec(x) = tan(x)sec(x)
dx
Z
sec(x) = tan(x)sec(x)
d
tan(x) = sec2 (x)
dx
Z
tan(x) = sec2 (x)
Starting with the formula and dividing by cos2 (x):
sin2 (x) + cos2 (x) = 1
)
tan2 (x) + 1 = sec2 (x)
This allows us to interchange tan2 (x) and sec2 (x).
Only 2 powers of sec(x)
Use substitution, set u = tan(x).
Z
sec2 (x)tan100 (x)dx
u = tan(x)
Z
2
sec (x)tan
100
(x)dx =
Z
du = sec2 (x)dx
u100 du =
u101
tan101 (x)
=
101
101
Only 1 power of tan(x) and at least 2 powers of sec(x)
Note that for 1 power of sec(x), the answer is above.
Use substitution, set u = sec(x).
Z
tan(x)sec100 (x)dx
Z
u = sec(x)
du = tan(x)sec(x)dx
Z
100
tan(x)sec (x)dx = tan(x)sec(x)sec99 (x)dx
=
Author: Xiao
Z
u99 du =
u100
sec100 (x)
=
100
100
Page 83
MAT137
27 TRIGNOMETRIC INTEGRALS
Even powers of sec(x)
Use the formula tan2 (x) + 1 = sec2 (x) to interchange sec2 (x) with tan2 (x)
so that only 2 powers of sec(x) remain.
Z
Z
4
100
sec (x)tan (x) = sec2 (x)sec2 (x)tan100 (x)
Z
= sec2 (x)(tan2 (x) + 1)tan100 (x)
Z
Z
2
102
= sec (x)tan (x) + sec2 (x)tan100 (x)
Both are done with the substitution u = tan(x).
Odd powers of tan(x)
Replace so that only 1 power of tan(x) remain.
Z
Z
5
100
tan (x)sec (x) = tan(x)tan4 (x)sec100 (x)
Z
2
= tan(x) sec2 (x) 1 sec100 (x)
Z
= tan(x) sec4 (x) 2sec2 (x) + 1 sec100 (x)
Z
Z
Z
104
102
= tan(x)sec (x) 2 tan(x)sec (x) + tan(x)sec100 (x)
All are done with the substitution u = sec(x).
Note: Odd powers of sec(x) and even powers of tan(x) are very difficult.
Z
sec(x) = ln |sec(x) + tan(x)|
Multiples of csc(x) and cot(x)
Use 1+cot2 (x) = csc2 (x) and csc0 (x) =
cot(x)csc(x) and cot0 (x) =
csc2 (x)
Di↵erent frequencies of sin(mx) and cos(nx)
1
sinAcosB = [sin(A B) + sin(A + B)]
2
1
sinAsinB = [cos(A B) cos(A + B)]
2
1
cosAcosB = [cos(A B) + cos(A + B)]
2
Author: Xiao
Page 84
MAT137
27 TRIGNOMETRIC INTEGRALS
Question 77
Z
1.
sin4 x cos x dx
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
5
3
3
2
sin x cos x dx
sin x cos x dx
3
2
cos x dx
14.
16.
2
sin x dx
17.
2
sin x cos x dx
4
sin x dx
18.
19.
2
sec x dx
20.
tan x sec x dx
5
13.
15.
cos x dx
2
12.
2
tan x sec x dx
21.
22.
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
tan x sec4 x, dx
sec4 x tan4 x dx
tan3 x sec x dx
sec4 x dx
tan3 x dx
sec5 x dx
tan2 x dx
tan2 x sec x dx
sin 2x sin 3x dx
cos 2x sin 5x dx
cos2 3x sin 2x dx
Question 78
Verify the Orthogonal Relation for Fourier Sine and Cosine Series:
(
(
Z ⇡
Z ⇡
0
if m 6= n
0
if m 6= n
sin mx sin nx dx =
cos mx cos nx dx =
⇡/2 if m = n
⇡/2 if m = n
0
0
Z ⇡
sin mx cos nx dx = 0 (for any n, m)
where n, m are integers.
⇡
Author: Xiao
Page 85
MAT137
28
28 TRIGNOMETRIC SUBSTITUTION
Trignometric Substitution
p
Integral involving pa x2
This looks similar to 1 sin2p
(✓) = cos(✓)
Try to turn the expression into 1 z 2 for some new z.
Use substitution, set z = sin(✓).
Z
x2
p
dx
4 x2
s ✓
r
r
◆
2
2
p
4x
x
4 x2 = 4
= 4 1
=2 1
4
4
x
= sin(✓)
2
dx
= 2cos(✓)
d✓
Z
Z
x2
x2
p
q
dx =
4 x2
2 1
)
⇣ x ⌘2
2
x = 2sin(✓)
)
x
2
dx = 2cos(✓)d✓
Z
x2
p
dx =
· 2cos(✓)d✓
2
2 1 sin(✓)2
Notice we still have x in the expression, so we need to change them to ✓.
Z
Z
4sin2 (✓)
p
=
· 2cos(✓)d✓ = 4sin2 (✓)d✓ = 2✓ sin(2✓)
2
2 1 sin(✓)
Now we need to express ✓ in terms of x.
For the first term, we can do ✓ = arcsin(x/2).
However, the second term involves a sin(2✓). While substituting the above
✓ expression gives the right answer, we need to simplify the answer.
The substitution relates the triangle with the
p opposite side length x and
hypotenuse length 2. The adjacent length is 4 x2 .
Using the double angle formula,
p sin(✓ + ✓) = sin(2✓) = 2sin(✓)cos(✓).
Using the triangle, cos(✓) = 4 x2 /2.
p
⇣x⌘
⇣x⌘
4 x2
= 2arcsin
2sin(✓)cos(✓) = 2arcsin
x
2
2
2
Author: Xiao
Page 86
MAT137
28 TRIGNOMETRIC SUBSTITUTION
p
Integral involving pa + x2
This looks similar to 1 + tan2p
(✓) = sec(✓).
Try to turn the expression into 1 + z 2 for some new z.
Use subtitution, set z = tan(✓).
p
Integral involving px2 a
This looks similar to sec2 (✓) p1 = tan(✓).
Try to turn the expression into z 2 1 for some new z.
Use substitution, set z = sec(✓).
p
Integral involving ax2 + bx + c
Complete the square inside the square root.
Then, depending on the situation, do a trig substitution from above.
Z
1
p
dx
(x + 1) x2 + 2x 3
Recall the procedure for completing the square: add and subtract (b/2)2 .
x2 + 2x
Z
3 = x2 + 2x
3 + 1 1 = (x + 1)2 4
Z
1
p
dx =
dx
3
(x + 1) (x + 1)2 4
1
(x + 1) x2 + 2x
p
This looks similar to sec2 (✓) 1. Changing the 4 to 1:
Z
Z
1
1
q
q
dx =
2
(x + 1)2 ( x+1
)2
(x + 1) 4( (x+1)
1)
2
4
p
x+1
= sec(✓)
2
)
x = 2sec(✓)
dx
1
1
dx
= 2sec(✓)tan(✓)
)
dx = 2sec(✓)tan(✓)d✓
d✓
✓
◆
Z
Z
1
sec(✓)
✓
1
x+1
=
·2sec(✓)tan(✓)d✓ =
d✓ = = arcsec
(x + 1)2tan(✓)
2sec(✓)
2
2
2
Question
Z p 792
Z
Z
9 x
1
1
p
1.
dx
2.
dx
3.
x4
(x2 + 2)3/2
ex e2x
Author: Xiao
4
dx 4.
Z
p
x2
1
2x
Page 87
3
dx
MAT137
29
29 PARTIAL FRACTION
Partial Fraction
We want to integrate any rational function,
f (x) =
p(x)
q(x)
where p(x), q(x) are polynomials.
We use a process that together is typically referred to as partial fractions.
Long Division
Long division is the first step in the process of integrating rational function.
We compare the degrees of the numerator p(x) and the denominator q(x).
(Degree of a polynomial p(x) is the value of its highest power.)
If deg p(x) < deg q(x), then we skip long division and proceed to the next step.
If deg p(x)
deg q(x), we perform long division on
p(x)
, and produce
q(x)
p(x)
r(x)
= g(x) +
q(x)
q(x)
where g(x) is the quotient polynomial, r(x) is the remainder polynomial.
p(x)
, we integrate the 2 terms on the right.
q(x)
The polynomial g(x) is easy to integrate.
r(x)
We only need to integrate the rational function
.
q(x)
It is guaranteed that deg r(x) < deg q(x), and thus we have made sure the
numerator has degree less than the denominator.
To integrate
Thus using the technique of long division, we can reduce any integral of
a rational function into a rational function with degree of the top less than
the degree of the bottom.
From now on, we will assume deg p(x) < deg q(x).
Author: Xiao
Page 88
MAT137
29 PARTIAL FRACTION
Partial Fraction
1. Completely factor the denominator q(x) into irreducible factors.
The irreducible factors must have either degree 1 or degree 2.
q(x) = q1 (x) . . . qn (x)
Let’s first assume each qi (x) are distinct (di↵erent from each other).
2. Write the original fraction into a sum of fractions of polynomials, with
qi (x) as denominators.
p(x)
p1 (x)
pn (x)
=
+·+
q(x)
q1 (x)
qn (x)
If qi (x) is degree 1, we set pi (x) to be a constant, such as pi (x) = A.
If qi (x) is degree 2, we set pi (x) to be degree 1, such as pi (x) = Ax + B.
Example
x3 + x2 x + 1
A
B
Cx + D
= +
+ 2
2
x(x 1)(x + 1)
x x 1
x +1
3. Take a common denominator on the right side. Solve for the constants
in pi (x) based on p(x) as a system of linear equations.
Example
x3 + x2 x + 1
A (x
= ·
2
x(x 1)(x + 1)
x (x
1)(x2 + 1)
B x(x2 + 1) (Cx + D) x(x
+
·
+ 2
·
1)(x2 + 1) x 1 x(x2 + 1)
x + 1 x(x
x3 + x2 x + 1 = A(x 1)(x2 + 1) + Bx(x2 + 1) + (Cx + D)x(x 1)
Expand and factor out xn :
x3 +x2 x+1 = (A+B +C)x3 +( A C +D)x2 +(A+B D)x A·1
Compare the coefficients in front of the powers of xn , we see that
1=A+B+C
1= A C +D
1=A+B D
1= A
Solving this system, we see that
A = 1, B = 1, C = 1, D = 1.
So we attain
x3 + x2 x + 1
1
1
x+1
=
+
+ 2
2
x(x 1)(x + 1)
x
x 1 x +1
Author: Xiao
Page 89
1)
1)
MAT137
29 PARTIAL FRACTION
4. In the event that a factor has multiplicity n, that is to say q(x) have
n
a repeated irreducible factor, such as q1 (x) ,
n
q(x) = q1 (x) q2 (x) . . . qn (x)
then to form a sum of fractions for this factor, we must repeatedly form
k
new fractions in the sum, with denominator q1 (x) with k increasing
from 1 up to n.
The numerator is still determined by whether q1 (x) is linear or quadratic.
Example
p(x)
3x2 + x + 1
A
B
C
D
=
= +
+
+
3
2
q(x)
x(2x 1)
x 2x 1 (2x 1)
(2x 1)3
Notice that we have (2x 1)k as the denominator, with k = 1, 2, 3.
In each of the instances, we have a constant as numerator, since (2x 1)
as the repeated factor, is linear.
Example
p(x)
3x2 + x + 1
A B
Cx + D
Ex + F
Gx + H
= 2 2
= + 2+ 2
+ 2
+ 2
3
2
q(x)
x (x + 1)
x x
(x + 1) (x + 1)
(x + 1)3
Notice that we have xk as denominator, with k = 1, 2.
The numerators are set to be constants, since the irreducible repeated
factor x is linear. That is to say, we view x2 = (x)2 .
However, x2 + 1 is an irreducible deg 2 repeated factor, so we have
(x2 + 1)k as denominator, with k = 1, 2, 3.
The numerators are set to be linear, since the irreducible repeated factor (x2 + 1) is quadratic.
Thus, the partial fraction method writes f (x) as a sum of fractions with deg
1 or deg 2 denominator, and each fraction can be integrated separately.
We are left with solving integrals of the following forms:
Z
Z
A
Ax + B
k
2
(ax + b)
(ax + bx + c)k
where ax2 + bx + c is an irreducible quadratic.
Author: Xiao
Page 90
MAT137
29 PARTIAL FRACTION
Integration
1. Linear term (degree 1) on denominator.
Z
A
(ax + b)k
take substitution u = ax + b, to result in an integral of the form
Z
1
uk
2. Quadratic term (degree 2) on denominator.
Linear term on numerator.
Z
Ax + B
2
(ax + bx + c)k
We can split the fraction using +.
We first focus on the term with x in the numerator.
Z
Ax
dx
2
(ax + bx + c)k
We must set u = ax2 + bx + c from the denominator.
Then du = (2ax + b)dx as a linear term.
So we try to make the numerator to be du = (2ax + b)dx exactly.
Z
Z
Z
⌘
A
2ax
A⇣
2ax + b
b
=
=
dx+
2a
(ax2 + bx + c)k
2a
(ax2 + bx + c)k
(ax2 + bx + c)k
The first integral after the substitution u = ax2 + bx + c result in
Z
1
uk
The second integral is of the form
Z
Z
B
1
=B
2
k
2
(ax + bx + c)
(ax + bx + c)k
which is the form of the 2nd term in the fraction.
Author: Xiao
Page 91
MAT137
29 PARTIAL FRACTION
3. Quadratic term (degree 2) on denominator.
Constant term (equivalent to the number 1) on the numerator.
Z
1
2
(ax + bx + c)k
This case is difficult, we focus on the case of k = 1.
Z
1
2
ax + bx + c
By the partial fraction procedure, the quadratic must be irreducible.
We complete the square on the denominator, and try to attain the
form
Z
1
⇣
⌘2
cx + d + 1
We take u = cx + d.
This gives the integral of the form
Z
1
= arctan(u)
2
u +1
The case of k > 1 requires integration by part to reduce to the case of k = 1.
In conclusion, the result of the integral
Z
p(x)
q(x)
only involve terms of the form: um , ln(u), arctan(u).
(where m can be positive or negative)
Question 80
1.
2.
Z
Z
2x3 + x2 x
(x 1)2
4
dx
x3 + x2 + x + 2
dx
x4 + 3x2 + 2
Author: Xiao
3.
4.
Z
Z
x3
x+1
dx
2x2 + 2x
(x2
1
dx
+ 1)2
Page 92
MAT137
30
30 SEQUENCES
Sequences
A sequence S = {an }1
n=1 is like a regular function f (n) = an , except you can
only ask its limit as n ! 1.
A sequence S is bounded above,
if every number in the sequence is less than some big number M.
If a sequence S is bounded above, there exists a quantity that is very similar
to the maximum of S, it is called the supremum of S, denoted sup S.
It is like the maximum, because, if you take a step to the left, from the
supremum, no matter how small the step is, you will bump into something in S along the way.
8✏ > 0, 9s 2 S such that, sup S
✏ < s  sup S
Similarly, if S is bounded below, then there exists a quantity that is similar
to the minimum of S, called infimum of S, denoted inf S.
Of course, you take a step to the right, from the infimum, you will bump
into something in S.
8✏ > 0, 9s 2 S such that, inf S  s < inf S + ✏
The definition of limit as n ! 1 for S = {an } and the computation of limit
is exactly the same as a regular function f (n) = an .
For a sequence {an } , the limit L exists when the values of an get close
to L. How close is close?
For any error range ✏, as long as we take n large, we can always make sure
an is within that error to L
8✏ > 0, 9N such that, if n > N ) |an
L| < ✏
If the limit exists, then S gets close to L, so it is not going to infinity, so it’s
bounded above and below. However, being bounded (above and below) is
not enough for the limit to exist.
Author: Xiao
Page 93
MAT137
30 SEQUENCES
If the limit of S = {an } does not exist, we say that it diverges.
It coule be going to infinity, or it could oscillate and not approach a specific number L.
If S is unbounded, S would get bigger and bigger, so it can’t converge.
increasing means each term is bigger than the term before: 8n, an+1 > an .
non-decreasing is just a technical way of saying 8n, an+1 an .
If S is bounded above, and it’s non-decreasing, then it converges to its
’maximum’ value, or the supremum.
decreasing means 8n, an+1 < an , non-increasing means 8n, an+1  an .
If S is bounded below, and it’s non-increasing, then it converges to its
’minimum’ value, or the infimum.
Being non-increasing or non-decreasing is called monotone.
An important note: the limit of a sequence has nothing to do with how
it behaves at the beginning. We only ask about how the sequence behaves
as n ! 1, so we can ignore finitely many terms when computing limit.
Using this, we can say the following are true ’eventually’, which means when
n is large enough:
ln(n) ⌧ np
for any power of p > 0. (log grows slower than polynomial)
n p ⌧ an
for any base a > 1. (polynomial grows slower than exponential)
an ⌧ n!
for any base a > 1. (factorial grows very fast)
n
n! ⌧ n .
(nn grows the fastest)
We attain the chain of ascending growth rates:
p
1 ⌧ ln(n) ⌧ n ⌧ n ⌧ n2 ⌧ ... ⌧ 2n ⌧ en ⌧ n! ⌧ nn
where an ⌧ bn if
an
=0
n!1 bn
Compute limits when the base of not a constant
To compute limit, L’Hopital Rule is usually the method of choice.
Change to base e, then bring the limit to the exponent:
lim
a = eln(a)
Author: Xiao
ak = ek·ln(a)
Page 94
MAT137
30 SEQUENCES
Example
(n + 2)1/n = eln(n+2)
1/n
= eln(n+2)/n
Using the chain of growth, the exponent approach zero, thus the answer to
the limit is e0 = 1.
Question 81
Find the infimum and supremum of the following sets.
1. S = {x : |x
1| < 2}
2. S = {0.9, 0.99, 0.999, ....}
3. S = {( 1)n (1
1
) : n 2 N}
n
Question 82
Find the limit of the following sequences.
n + ( 1)n
n
6.
1
2n
4n
3. p
4n2 + 2
7.
n
ln n
2n
3n + 5
n+1
5. n p
2 n
8. (n + 2)1/n
1.
2. (1.01)n
4.
1
2n + 3
3n+1
9. n 1
4
Question 83
Prove an ! 0 if and only if |an | ! 0.
Question 84
Consider the recursively defined sequence.
Determine whether it converges. If it converges, find its limit.
1
1. a1 = 1, an+1 = an + 1.
2
p
p
2. a1 = 2, an+1 = 2 + an
Author: Xiao
Page 95
MAT137
31
31 IMPROPER INTEGRAL
Improper Integral
Integral is many rectangle’s area which is base times height.
What happens if one of them is infinite?
Type 1: base is infinite
This corresponds to one of the limit of integration being infinity.
Z 1
Z b
f (x) = lim
f (x)
a
b!1
a
We integrate to some big number b as usual, then take the limit.
Type 2: height is infinite
This corresponds to the function going to infinity.
Assume f (x) appraoches infinity at x = k, and a < k, then
Z k
Z b
f (x) = lim
f (x)
a
b!k
a
We integrate to some number b close to k, then take the limit.
When multiple infinite points arise
We need to split the integral into several pieces, so that in each piece, only
1 of the 2 bound is approaching an infinite point.
When one piece results in infinity, the entire integral diverges.
Example
Z 1
1
(x 1)3
0
The integral has infinite points at x = 1, and x = 1. We need to approach
x = 1 on both sides. So we choose some number c 2 (1, 1) and split up the
integral:
Z 1
Z c
Z 1
1
1
1
=
+
+
3
3
1)
1)
(x 1)3
0 (x
1 (x
c
Z b
Z c
Z b
1
1
1
= lim
+ lim+
+ lim
3
3
b!1 c (x
b!1
b!1
1)
1)
1)3
0 (x
b (x
1
= 1+1+
= ±1
2(c 1)2
Author: Xiao
Page 96
MAT137
31 IMPROPER INTEGRAL
Integrals can’t always be computed.
However, we can ask whether an integral is convergent (not infinite).
Comparison Test for Improper Integrals
If f, g continuous, and 0  f (x)  g(x) for x a, Then
Z 1
Z 1
0
f (x) 
g(x)
a
a
In particular:
R1
R1
If the larger integral aR g(x) < 1, then the smaller integral a f (x)
R 1< 1.
1
If the smaller integral a f (x) = 1, then the larger integral 1  a g(x).
To prove an integral converges, compare it to a larger integral which can
be computed to converge.
To prove an integral diverges, compare it to a smaller integral which can be
computed to diverge.
Typically, the new integral will be modified from the original, by removing complicated terms.
Limit Comparison Test for Improper Integrals
If f, g continuous, both positive for x a, If
f (x)
6= 0 6= 1
x!1 g(x)
lim
Then
f (x) ⇡ g(x)
In particular:
Z 1
a
f (x) converges if and only if
(The rate of growth is similar)
Z
1
g(x) converges
a
This is typically used on rational functions.
When f (x) is a rational function, we take the leading powers on the top
and bottom to create a new function g(x).
This method guarantees the the limit is not zero or infinite.
Author: Xiao
Page 97
MAT137
31 IMPROPER INTEGRAL
Question 85
Compute the improper integral, or show it diverges.
Z 1
Z 1
1
1.
5.
cos x
2
1 x +1
0
Z 1
Z 1
1
1
2.
6.
2
x+2
2
0 x
Z 1
Z 1
1
p
3.
ln x
7.
x
0
0
Z 1
Z 1
1
4.
ln x
8.
2
x
1
1
2
Question 86
Let p > 0. For which values of p would the integral converge or diverge?
Z 1
1
xp
1
This integral is sometimes called the p-series.
Question 87
Using the comparison tests, show whether the integral converges or diverges.
Z 1
sin x + 2 cos x + 10
1.
x2
1
Z 1
x 7
2.
2
x +x+5
0
p
Z 1
x 6
3.
2
10 3x + 5x + 11
Z 1
arctan x
4.
x1.1
0
Z 1
sin x
5.
4/3
0 x
Z 1
2
6.
e x
0
Author: Xiao
Page 98
MAT137
32
32 SERIES
Series
We would like to add numbers together, but it gets tricky if we want to add
infinitely many of them.
We define a series as a sequence of partial sums.
For a series
1
X
an
n=1
Consider the sequence:
s1
s2
s3
s4
= a1
= a1 + a2
= a1 + a2 + a3
= a1 + a2 + a3 + a4
..
.
sn = a1 + a2 + a3 + ... + an
The sequence {sn } are the partial sums.
We define the series converge if the sequence {sn } converge.
Usually, infinite series are very difficult to calculate. However, one important
result is the geometric series, which converges:
1
X
xn =
n=0
1
1
|x| < 1
x
Only when series converge, we can perform regular algebra as follows:
1
X
an +
n=a
1
X
bn =
n=a
c·
1
X
n=a
an =
1
X
an + b n
n=a
1
X
n=a
c · an
With these, we can manipulate the geometric series to attain exact answers
to some series.
Author: Xiao
Page 99
MAT137
32 SERIES
Convergence of Series
Most of the time, we can not expect to attain the exact answer to the series.
Instead, we can characterize whether a series converge or not.
When taking this approach, we again arrive at the previous conclusion that:
we only care about how the terms behave as n ! 1, and we can ignore
finitely many terms.
Thus, the starting point of the series does not a↵ect convergence.
However, since we can not ignore infinitely many terms, if the terms get
larger and larger, the series can’t possibly converge:
Test 0: Zero Test for Divergence
1
X
If lim an 6= 0, then
an diverges.
n!1
n=a
However, even when lim an = 0, it might oscillate, or diverge for other reasons.
n!1
Thus test 0 is almost never useful.
When the series does not oscillate, it usually will be decreasing ’eventually’, and will remain positive, then we can connect the points an smoothly
to form a continuous function.
The series can be thought of as area of the rectangles below the curve, so it
must be the case that the integral is approximately equal to the series.
Test 1: Integral Test
For an positive and decreasing ’eventually’,
1
X
n=a
an ⇡
Z
1
an
a
In particular, the sum (series) converges if and only if the integral converges.
We should all know by now that integrals can get really difficult. Although
this approximation can almost always be made, it is almost never useful.
(Decreasing ”eventually”: there exists N such that for all n > N , an+1 < an .)
Author: Xiao
Page 100
MAT137
32 SERIES
P-series
Which integrals are easy to do? Clearly powers like xp .
For a power p:
Z 1
Z 1
Z 1
1
X
1
1
1
⇡
=
dx =
x p dx
p
p
p
n
n
x
a
a
a
n=a
For p 6= 1, this is easily calculated:
x p+1
p+1
=
x=1
x=a
The evaluation at x = a is just a number so it’s irrelevant, and so we only
need to focus on the infinity part.
The answer depends upon the exponent. If the exponent is positive, then it’s
infinite. If the exponent is negative, then it’s finite. For p = 1, then answer
is ln(x), which is infinite. Thus:
Z 1
1
X
1
1
⇡
dx
p
n
xp
Diverges for p  1
Converges for p > 1
Given this result, what can we say about:
1
X
1
n2
1
Well, it most likely converges because we only care about the behaviour as
n becomes large, so the number 1 is irrelevant.
In fact, this is true in general:
When an is a rational function, we only need to take the leading powers
1
X
on the top and bottom to create a new series
bn .
Test 2: Limit Comparison Test
For an , bn positive ’eventually’,
1
1
X
X
an
6= 0 6= 1, then
an ⇡
bn
n!1 bn
If lim
In particular, the first sum (series) converges if and only if the second converges.
Author: Xiao
Page 101
MAT137
32 SERIES
Now that we are able to test convergence for rational function, we next need
to worry about other functions, such as ln(x), sin(x), cos(x), arctan(x).
Fortunately, a function like sin(x) is at most 1, so it being inside a series
would at most be multiplying the original series by 1, which shouldn’t matter
if the series already converge.
Test 3: Comparison Test
Suppose an  bn and positive ’eventually’, then
If
1
X
If
bn Converges, then
1
X
an Diverges, then
1
X
1
X
1
X
an 
1
X
bn :
an Converges
bn Diverges
Note that in both cases, the implication is only one direction.
1
X
arctan(n)
n2
So the left series converges too.
1
X
⇡ 1

Converges
2 n2
Using ln(n) < np for any p > 0:
1
X
ln(n)
1 p
X
n
<
n2
So the left series converges too.
1
X
1
=
Converges
2
3/2
n
n
Now we focus on series including exponents an and factorials n!.
Test 4: Ratio Test
For an positive ’eventually’, compute
an+1
lim
=L
n!1 an
If L < 1, then
1
X
an converge.
1
X
If L > 1, then
an diverge.
If L = 1, then it’s inconclusive.
Author: Xiao
Page 102
MAT137
32 SERIES
Convergence of Series involving Positive and Negative Terms
In general, this is a difficult problem.
Thus it’s much easier to just get rid of all the negative signs.
Test 5: Absolute Convergence Test
1
1
X
X
If
|an | Converges, then
an Converges
The quantity |an | means making every term positive.
Intuitively, if the positive series converge, then some terms becoming negative
on the right would only make the sum smaller, and would still converge.
1
1
X
X
When
|an | converges, we say
an is absolutely convergent.
Note: the positive sum diverging does not mean the original series diverge.
1
1
X
X
In the case when the
an does converge but
|an | does not converge,
1
X
we say
an is conditionally convergent.
1
X
sin(n)
n2
This series oscillate due to sin(n), but putting on absolute values:
1
1
X
X
sin(n)
1
|
|
Converges
2
n
n2
Thus the original series converges absolutely by comparison test.
A special case is when the positive and negative terms alternate in turn:
a1 a2 + a3 a4 ... This can be written with the notation ( 1)n or ( 1)n+1 .
Test 6: Alternating Series Test
For an positive and decreasing ’eventually’, consider the alternating series
1
X
( 1)n an = a1 a2 + a3 a4 ....
Then we only need to compute the limit of an :
1
X
lim an = 0, if and only if
( 1)n an Converges
n!1
The test also holds with cos(n⇡) = ( 1)n = sin (n + 1/2)⇡ , and ( 1)n+1 .
Author: Xiao
Page 103
MAT137
33
33 SUMMARY OF TESTS FOR SERIES
Summary of Tests for Series
Combining all the tests, the steps for determining convergence of a series
1. Take the absolute value of an
1
X
an :
2. If there are exponents (2n ) or factorial (n!), apply Ratio Test (or Root Test):
⇣
⌘
p
an+1
lim |
|=L
or lim n |an | = L
n!1 an
n!1
1
X
L < 1 conclude
an absolutely converge.
1
X
L > 1 conclude
an diverge.
3. ln(n) in the numerator, compare 1 < ln(n) < np with p small.
ln(n) in the denominator, compare 1 > 1/ln(n) > 1/np with p small.
For sin and cos in numerator, compare | sin(n)|  1 ,| cos(n)|  1.
For arctan in numerator, compare arctan(n) < ⇡/2.
4. For series in rational function form, take bn to be leading powers of
top and bottom. Attain bn = n1p .
1
X
p > 1 conclude
an absolutely converge.
1
1
X
X
p1
|an | diverge, but
an may still converge.
1
X
5. If the original series can be written in the form
( 1)n an .
For an decreasing, compute lim an = L .
n!1
1
X
L = 0 conclude
( 1)n an conditionally converge.
1
X
L 6= 0 conclude
( 1)n an diverge.
1
, there is one power of n on the bottom, use integral test.
n · ln(n)
Use small angle approximation on:
For an =
sin(1/n) ⇡ 1/n ⇡ tan(1/n) ⇡ arcsin(1/n) ⇡ arctan(1/n)
1
X
If lim an 6= 0, then conclude
an diverge (by Test 0).
n!1
Author: Xiao
Page 104
MAT137
33 SUMMARY OF TESTS FOR SERIES
Question 88
Show whether the series is Absolutely convergent, Conditionally convergent, or Divergent.
1.
1
X
n=1
2.
1
X
n=2
3.
4.
1
2 + cos n
13.
n=1
15.
1
X
1
n2
n=1
16.
17.
n2 + 4n
p
6.
n5 + 4n4 + 2n
n=1
1 p
X
n sin n
7.
n2
n=1
( 1)n sin
n=1
1
X
1
n
1
9.
( 1) sin 2
n
n=1
11.
12.
n
1
X
(n + 1)2n
n=1
1
X
cos n⇡
n!
1
X
n!
23n
n=1
1
X
n!
(2n)!
n=1
Author: Xiao
ln n
1
X
ln n
n=2
1
X
10.
1
X
( 1)n 2n
nn
n=1
n=2
1
X
( 1)n
5.
n2 1
n=2
8.
(3n)!
1
X
1
14.
nn
n=1
1
n ln n
1
X
( 1)n
n
n=1
1
X
1
X
n!(2n)!
n3
1 p
X
n ln n
18.
(n + 1)2
n=2
1
X
p
1
X
n2 (ln n)2
1
n ln n
n=2
p
1
X
n+1
20.
1.4
n (ln n)2
n=2
19.
21.
n=2
1
1
X
(ln n)10
p
22.
n
n=2
1
X
1
n(ln n)3
n=2
p
1
X
n4 + 2n3 + 1
p
24.
3
2n9 + 4n8 + 5n6 (ln n)5
n=2
23.
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MAT137
34
34 TAYLOR SERIES
Taylor Series
We would like to approximate the function f (x) with a polynomial.
We choose the point a as the center of expansion.
We require the polynomial to have the same derivatives as the function at a.
This (infinite) polynomial is called the Taylor (power) Series of f (x):
1
X
f (n) (a)
f (x) ⇡ P (x) =
(x a)n
n!
n=0
Usually, the approximation only holds in a symmetric interval centered at a.
This is called the interval of convergence.
Half of the length of the interval is called the radius of convergence.
an+1
Use Ratio Test to attain the interval of convergence, set |
| < 1.
an
The series may converge or diverge on the boundary of the interval.
For a = 0, the most important Taylor Series to know are:
1
X
xn
x2 x3
x
e =
=1+x+
+
+ ...
x2R
n!
2
3!
n=0
sin(x) =
1
X
( 1)n 2n+1
x
=x
(2n
+
1)!
n=0
cos(x) =
1
1
x
1
X
( 1)n 2n
x =1
(2n)!
n=0
=
1
X
x3 x5
+
3!
5!
x2 x4
+
2!
4!
x2R
...
...
xn = 1 + x + x2 + x3 + ...
n=0
x2R
|x| < 1
(The note x 2 R means the approximation holds for any number x.)
Since the formula holds for any x inside the interval of convergence, we can
substitute x with any other quantity.
Example
e
x2
=
1
X
(x2 )n
n=0
1
3
n!
= 1 + x2 +
x4 x6
+
+ ...
2
3!
x2R
i
1
1
1 X x n 1h x x 2 x 3
= ·
=
( ) = 1+ +( ) +( ) +...
x
3 1 x/3
3 n=0 3
3
3
3
3
Author: Xiao
1
x
<1
3
Page 106
MAT137
34 TAYLOR SERIES
Derivatives and Integrals
We can take derivatives and integrals of Taylor Series on both sides of the equation.
For the sum, we take derivatives and integrals term by term. In other words,
Z ⇣X ⌘
XZ
X d
d X
an dx =
an dx
an =
an
dx
dx
The interval of convergence remains the same.
Example: arctan(x) and ln(1-x)
1
1
X
X
1
1
2 n
=
=
( x) =
( 1)n x2n
1 + x2
1 ( x2 ) n=0
n=0
|
x2 | < 1
(We get | x2 | < 1 since we plugged in x2 , but this is equivalent to |x| < 1.)
Taking integral on both sides:
Z
1 Z
X
1
= arctan(x) = C +
( 1)n x2n
|x| < 1
2
1+x
n=0
arctan(x) = C +
1
X
( 1)n
n=0
To solve for C, plug in x = 0, x
arctan(x) =
2n+1
1
X
= 0, so arctan(0) = 0 = C +
( 1)n
n=0
Similarly,
ln(1
x) =
1
X
xn
n=1
n
=
x2n+1
2n + 1
x2n+1
2n + 1
x2
2
x
|x| < 1
x3
3
...
P
0.
|x| < 1
Limit Using Taylor Series for x ! 0
We can plug in the Taylor Series of the common functions into the limit.
The higher powers of the Taylor Series would be zero after taking x ! 0,
leaving the answer as the leading term.We can divide top and bottom by
the lowest power to attain the leading term.
3
5
x x3! + x5! ...
sin(x)
x2 x4
lim
=
=1
+
...
x!0
x
x
3!
5!
Plugging in x = 0, we attain the answer 1.
Note: If the limit is x ! a, we need to expand the Taylor series at a.
Author: Xiao
Page 107
MAT137
34 TAYLOR SERIES
Question 89
Find the radius of convergence and the interval of convergence for the power series.
1.
1
X
n(x
2)n
n=1
2.
1
X
x3n
2n
n=2
3.
1
X
( 1)n (x
3)2n
n=3
1
X
(n!)2
4.
(x
(2n)!
n=4
3)n
Question 90
Find the Taylor Series of the functions centered at a = 0.
1.
1
1+x
1
2.
1 + x2
2 x
3. x e
4. cos(x4 )
5. sin(2x)
6. e
x
7. ex
e
8.
1
3x + 2
9.
x2
x
sinx
x
10. ln(1
x)
11. ln(1 + x)
Z
12. f (x) =
x
e
t2
dt
0
Question 91
Find the Taylor Series of the functions centered at the given value of a.
1. ex at a = 1
1
2.
at a = 3
1 x
⇡
3. sin(x) at a =
4
Author: Xiao
Page 108
MAT137
34 TAYLOR SERIES
Question 92
Find the exact value of the series (using Taylor Series).
1.
1
X
nx
n
5.
n=1
2.
3.
6.
2n
7.
n!
1
X
n=0
1
X
(n + 1)xn
8.
n!
n=0
1
X
xn
(n + 2)n!
( 1)n
n=0
1
X
xn+1
n=0
4.
n=0
1
X
n2
n=0
1
X
(n + 1)x2n+1
(2n + 1)!
xn
(n + 1)(n + 2)
1
X
(4n2 + 8n + 3)2n
n!
n=2
Question 93
View the infinite repeating decimal expansion as a series, and compute its value.
a) 0.99999...
c) 0.252525...
b) 0.11111...
d) 0.3121212...
Question 94
Find the limit using Taylor Series.
6 sin x 6x + x3
1. lim
x!0
x5
2. lim
x!0
ex
2
cos(2x) 3x2
x2 sin(x2 )
Question 95
A telescoping series is when adjacent positive and negative terms cancel,
N
X
so the N-th partial sum:
an would only consist of the first and the last
n=1
term, and we can compute the series by taking the limit N ! 1.
1.
1
X
[arctan n
n=0
⇣ n ⌘
2.
ln
n+1
n=1
1
X
Author: Xiao
arctan(n + 1)]
3.
1
X
n=1
1
n2 + 3n
1
X
n+2
4.
n3 n
n=3
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MAT137
35
35 POWER SERIES
Power Series
g(x) =
1
X
a)n
an (x
is called a power series.
n=0
For any power series g(x), there exists a unique radius of convergence R
1. The power series g(x) converges absolutely for |x
2. The power series g(x) diverges for |x
a| < R. (in the circle)
a| > R. (outside the circle).
If R > 0, then the power series g(x) (as a function of x) is analytic in
|x a| < R and g 0 (x) is given by term by term di↵erentiation
0
g (x) =
1
X
a)n
nan (x
1
n=1
with the same radius of convergence R.
Note that:
1. g(a) = a0 is always defined, and so the power series at least converge
at a. Thus R
0. However, g(x) may not converge for any other x,
and then we say R = 0.
2. If g(x) converges absolutely for all x, then we say R = 1, and so g(x)
is analytic everywhere.
3. g(x) may or may not converges on the boundary of the interval
(circle) of convergence, which needs to be checked separately.
4. The point a is called the center of expansion.
5. As g 0 (x) is also a power series, we have g(x) is in fact infinitely di↵erentiable in |x a| < R.
Given an analytic function f (x), then for any a in the domain, f (x) must
equal its Taylor Series in an interval (circle) near a:
f (x) =
1
X
f (n) (a)
n=0
n!
(x
a)n
Note that Taylor Series is a power series.
This is called expanding f (x) at the point a.
Author: Xiao
0:
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MAT137
35 POWER SERIES
The radius of convergence R can be obtained in 2 ways:
Ratio test:
R=
1
an+1
lim
n!1
an
Root Test:
R=
1
p
lim n |an |
n!1
Taylor Polynomial
If we only take finitely many terms of the Taylor Series, we get a polynomial
called the (N-th order) Taylor Polynomial PN (x). The Taylor Polynomial
will not be equal to f (x), but it will be an approximation. The di↵erence
between the Taylor Polynomial and f (x) is called the remainder RN (x).
PN (x) =
N
X
f (n) (a)
n=0
f (x) =
N
X
f (n) (a)
n=0
n!
(x
n!
a)n + RN (x)
(x
a)n
RN (x) =
f (n+1) (c)
(x
(n + 1)!
a)n+1
where c 2 (a, x) is undetermined by the mean value theorem.
The 1st-order Taylor Polynomial is the tangent line at a:
P1 (x) = f (a) + f 0 (a)(x
f (x)
x!a
x
The definition of the derivative, f 0 (a) = lim
a)
f (a)
gives:
a
f (x) P1 (x)
f (x) f (a) f 0 (a)(x a)
= lim
= f 0 (a) f 0 (a) = 0
x!a
x!a
x a
x a
As x approaches a, we know that x a is going to zero. The 1st order
approximation P1 (x) = f (a) + f 0 (a)(x a), as x approaches a, the error
(remainder) between the function f (x) and P1 (x) would also go to zero.
However, we have that this error (remainder) between the function f (x) and
the 1st order approximation is going to zero faster than 1 power of (x a).
lim
As N increaess, the speed for which (x a)N is going to zero increases.
In general, the N-th order Taylor Polynomial would have its error go to
zero faster than N powers of (x a):
lim
x!a
Author: Xiao
f (x)
(x
PN (x)
=0
a)N
Page 111