1
Sequences
To calculate the limit of a sequence an = f (n), you find limn→∞ f (n). The process for this
is basically the same as limits to infinity of functions of x, but you have to be careful about
things like an = sin(nπ), which have limits even though the corresponding functions of x do
not. In this case, limn→∞ sin(nπ) = 0 but limx→∞ sin(πx) DNE.
Exercise 1.1. Calculate the limit of an =
n2 +1
.
2n2 +n+1
Solution. You treat such rational functions as you would the corresponding functions of x:
by looking at the dominant terms. Since the dominant terms are n2 in the numerator and
2n2 in the denominator, we can say
n2
n2 + 1
1
∼ 2 = ,
2
2n + n + 1
2n
2
so the limit is 1/2. To do this precisely, we multiply the top and bottom by 1/n2 (1 over the
dominant term):
(n2 + 1) n12
1 + n12
=
.
(2n2 + n + 1) n12
2 + n1 + n12
As n → ∞, every term but the constants goes to 0, so we are left with 1/2, which is what
we got above as well.
Exercise 1.2. Calculate the limit of an = e−n cos(nπ).
Solution. We want limn→∞ e−n cos(nπ). Now, cos(nπ) may have no limit by itself, but just
like in the example in the beginning of the section, we should plug in values just to make
sure:
n = 0 : cos(0π) = 1,
n=1:
cos(π) = −1,
n=2:
cos(2π) = 1,
n=3:
cos(3π) = −1,
n
. The top just oscillates
and so on. So cos(nπ) is just (−1)n . So our sequence is an = (−1)
en
between ±1, and the bottom is tending towards ∞, so the quotient will tend to 0. Therefore
limn→∞ an = 0.
Remark. The above sequence behaves similar to the sequence an =
understand why this goes to 0.
√
Exercise 1.3. Calculate the limit of the sequence an =
(−1)n
.
n
Make sure you
n2 +1
.
n
Solution.
Again, we can look at behavior (since we are taking n → ∞). The top behaves
√
like n2 = |n| = n since n > 0, so the fraction behaves like n/n = 1. So the limit is 1. To
1
√
prove this “rigorously,” bring the n into the square root by writing n = n2 :
√
√
n2 + 1
n2 + 1
= lim √
lim
n→∞
n→∞
n
n2
r
n2 + 1
= lim
2
n→∞
r n
1
= lim 1 + 2 (divide the inside)
n→∞
n
√
1 = 1.
=
√
Remark. It is incredibly important to remember that x2 = |x|, not just x. This will be
needed later.
2
Limits and Continuity
Method 2.1 (Limits without L’Hopital). Remember that you had ways of calculating limits
before L’Hopital’s rule came along. Such tricks include factoring, multiplying by conjugates,
etc.
Definition 2.2. A function f (x) is continuous at x = a if:
(1) f (a) exists.
(2) limx→a f (x) exists.
(3) limx→a f (x) = f (a).
Remark. Notice that this definition allows us to pull limits in and out of continuous functions.
For example, if we wanted to calculate
lim e
sin(x)
x
x→0
,
since f (x) = ex is continuous, we can say
lim e
x→0
sin(x)
x
= elimx→0
sin(x)
x
= e1 = e.
Theorem 2.3 (Special Limits). We have:
sin(h)
lim
= 1,
h→0
h
eh − 1
lim
= 1,
h→0
h
1 − cos(h)
lim
= 0,
h→0
h
ah − 1
lim
= ln(a).
h→0
h
Exercise 2.4. Calculate the following limits without L’Hopital’s rule.
(a) limx→3
x2 −1
(x+1)(x+2)(x+3)
(b) limx→0
√
2
√x +1−1
x+4−2
2
(c) limx→2
x2 −4
x2 −5x+6
(d) limx→0
1−cos(x2 )
sin(2x2 )
(e) limx→1
2x−1 −1
x−1
Solution.
(a) To do this limit, just plug in 3, and you get
8
4·5·6
(b) You plug it in, you get 0/0, so we need to manipulate. Multiply by the conjugate of
both the top and bottom to get
√
√
x( x + 4 + 2)
x2 ( x + 4 + 2)
= lim √
= 0.
lim √
x→0 ( x2 + 1 + 1)
x→0 x( x2 + 1 + 1)
(c) Plug it in to get 0/0. Here we can factor the top and bottom:
x2 − 4
(x + 2)(x − 2)
x+2
=
lim
=
lim
= −4.
x→2 x2 − 5x + 6
x→2 (x − 2)(x − 3)
x→2 x − 3
lim
(d) Plug it in, get 0/0. To make it look like the sin(h)/h on bottom and (1 − cos(h))/h on
the top, multiply the top and bottom by 1/x2 :
1−cos(x2 )
x2
sin(2x2 )
x2
1−cos(x2 )
x2
lim 2 sin(2x
2)
x→0
2x2
1 − cos(x2 )
lim
= lim
x→0 sin(2x2 )
x→0
=
=
(multiply the top and bottom of the bottom fraction by 2)
0
= 0.
2
(e) When you plug it in you get 0/0. This looks like a special limit, but we have x → 1
instead of x → 0, so we make the substitution u = x − 1, so u → 0:
2x−1 − 1
2u − 1
= lim
= ln(2).
x→1 x − 1
u→0
u
lim
Exercise 2.5. Where are the following functions continuous:
(a) f (x) = ln(ex + 1)
q
(b) f (x) = ex1−1
(c) f (x) = sin(cos(x))
Solution. (a) Since ln(x) is continuous when x > 0, we need ex + 1 > 0. But ex is always
positive, so ex + 1 is as well. So this function is always continuous.
3
√
(b) Since x is continuous where x ≥ 0, we need ex1−1 ≥ 0. This will happen if and only
if ex − 1 > 0 (it can’t equal 0 since it is in the denominator). So we need ex > 1. This
means that x > 0 (look at the graph of ex to see this).
Remark. At the end, you could take ln of both sides to get x and 0, but to justify
that the inequality stays the same direction, you should say ln(x) is increasing (which
means it preserves inequalities).
(c) Since sin(x) and cos(x) are always continuous, this function is always continuous.
Exercise 2.6. Determine a so that
f (x) =
sin(ax)
x
2
2
a −x
if x < 0
if x ≥ 0
is continuous at x = 0.
Solution. Go through the 3 conditions in the definition:
(1) f (0) = a2 − 02 = a2
(2) To calculate the limits, take it from the left and right: The left limit is
lim−
x→0
sin(ax)
.
x
When you plug it in you get 0/0, so either use L’Hopital’s rule, or multiply the top
and bottom by a to use the special limit:
lim−
x→0
a sin(ax)
sin(ax)
= lim−
= a · 1 = a.
x→0
x
ax
The right limit is easier:
lim a2 − x2 = a2 − 0 = a2 .
x→0+
So for the limit to exist, we need a2 = a, so a2 − a = 0. Factoring: a(a − 1) = 0, so
a = 0 or a = 1.
(3) In both cases a = 0, 1, the limit matches the value of the function (which is a2 ), so
both make f (x) continuous.
3
Limits to Infinity
Limits to infinity are all about dominant terms. If the dominant term is on top, then the
limit is ±∞, if they’re the same, it’s the quotient of the coefficients, and if it’s on the bottom,
then the limit is 0.
Exercise 3.1. Calculate the following limits:
(a) limx→∞
2x2 −1
3x3 +x
4
(b) limx→∞
x+ex
2ex +x2
x3
x2 −1
limx→−∞ √xx2 +1
(c) limx→−∞
(d)
(e) limx→∞
x+cos(x)
x
Solution. (a) Here the dominant terms are 2x2 on top and 3x3 on the bottom, so the
bottom dominates. Therefore the limits is 0. To prove this properly, you do the same
trick as with sequences, except just multiply by the reciprocal of the dominant term
in the denominator (always).
2
(2x2 − 1) x13
− x13
x
=
lim
= 0.
x→∞ (3x3 + x) 13
x→∞ 3 + 12
x
x
lim
(b) Now the dominant terms are ex on the top and bottom, so the limit will be 1/2. Again
to prove this rigorously:
x
(x + ex ) e1x
+1
ex
=
lim
2 .
1
x→∞ (2ex + x2 ) x
x→∞ 2 + x
x
e
e
lim
To prove that the non constant terms go to 0, notice you have to use L’Hopital’s rule
since it is ∞/∞ for each. Make sure you can do this and get 0 for those, so the limit
is 1/2 as expected.
(c) The dominant term is on top, so the limit will be ±∞ (so it doesn’t exist). If you
needed to figure out which (if you were graphing this for example),
x3 · x12
x
lim
= lim
= −∞
x→−∞ 1 − 12
x→−∞ (x2 − 1) 12
x
x
since the top goes to −∞ and the bottom goes to 1.
√
(d) Here’s where√it’s important to know that x2 = |x|. The top is x, and the bottom
behaves like x2 = |x| = −x since x < 0 as x → −∞. So the fraction behaves like
x/ − x = −1.
(e) The dominant terms are the x’s since cos(x) is bounded between ±1, so the limit will
be 1. To prove this directly, break up the fraction:
x + cos(x)
cos(x)
= lim 1 +
.
x→∞
x→∞
x
x
lim
To show the second term goes to 0, we need Sandwich Theorem, which we will do next.
But the limit is 1.
5
4
Sandwich Theorem
Theorem 4.1 (Sandwich Theorem). If for x around c (except possibly at x = c), we have
g(x) ≤ f (x) ≤ h(x), and limx→c g(x) = limx→c h(x) = L, then limx→c f (x) = L.
Remark. One big clue that you want to use this theorem is when you have sin or cos of
something going to infinity, or sin or cos of 1/0.
Exercise 4.2. Calculate limx→0+ x sin(1/x).
Solution. If we plug in 0, we see the sin(1/0), so that is a clue for Sandwich Theorem. To
do this, we first bound the “problem,”, namely the sin(1/x):
−1 ≤ sin(1/x) ≤ 1.
We want x sin(1/x) though, so multiply through by x:
−x ≤ x sin(1/x) ≤ x.
Since the limit as x → 0+ of both sides is 0 (i.e. limx→0+ ±x = 0), the limit of our function
is 0, so
lim+ x sin(1/x) = 0.
x→0
Exercise 4.3. Calculate the following two limits:
(a) limx→∞
cos(x3 )
x2
(b) limx→π+ sin(x) cos sin
1
x−π
Solution. (a) First understand that this is a Sandwich theorem problem because of the
cos(x3 ) in the numerator (meaning there is a “cos(∞)”). Now that cos(x3 ) is the
problem in the limit, so we bound that:
−1 ≤ cos(x3 ) ≤ 1,
cos(x3 )
1
1
≤
≤ 2
2
2
x
x
x
2
(by just multiplying by 1/x everywhere). Now the limits of both ends of the inequality
are 0 since the top is constant and the bottom is getting larger. Therefore,
=⇒ −
cos(x3 )
= 0.
x→∞
x2
lim
(b) Again, first understand why this is a Sandwich Theorem problem. The first sin(x) isn’t
an issue, but the sin(1/0) inside the cos is a problem, and that sin(1/0) is the clue for
1
Sandwich Theorem. So we first bound the cos(sin( x−π
)) as
1
−1 ≤ cos sin
≤ 1,
x−π
6
because cos of ANYTHING is between −1 and 1. So
1
≤ sin(x).
− sin(x) ≤ sin(x) cos sin
x−π
The limit of ± sin(x) as x → π + is 0 (plug it in), so our limit is 0 as well.
1
Remark. You may have wondered by we didn’t bound just the sin( x−π
), i.e. why did we put
the whole cos expression. You could say
1
−1 ≤ sin
≤ 1.
x−π
The problem is if we need to take cos of everything, we cannot just assume the inequalities
stay the same direction, because cos(x) is neither always decreasing nor always increasing.
√
This is why we needed to bound that cos part as well. For functions like ex , ln(x), x,
etc. which we know are always either increasing or decreasing, we don’t need to worry. For
example, if we had
lim+ xesin(1/x) ,
x→0
we notice the sin(1/0) so we want to use Sandwich Theorem. However, we can just write
−1 ≤ sin(1/x) ≤ 1,
and then
e−1 ≤ esin(1/x) ≤ e1 ,
where we can justify the inequalities remaining the same because ex is increasing. So to
finish,
xe−1 ≤ xesin(1/x) ≤ xe1 ,
and since both ends go to 0, our limit is 0.
Exercise 4.4. If 1 −
x2
2
≤ g(x) ≤ 1 +
x4
4
for all x 6= 0, what is limx→0 g(x)?
Solution. Classic Sandwich Theorem: You have a function bounded by things and you want
to take limits. The limit of both ends are 1, so the middle limit is 1.
5
Intermediate Value Theorem and Bisection Method
Theorem 5.1 (Intermediate Value Theorem). If f (x) is continuous on [a, b] and L is a
number between f (a) and f (b), then there exists c in [a, b] with f (c) = L.
Method 5.2 (Bisection Method). Steps for bisection method: Start with interval
(1) Find the midpoint of your interval (in general, the midpoint of an interval [a, b] is
a+b
).
2
(2) Plug midpoint into the function
(3) Determine which half subinterval to pick by determining which half will satisfy the
conditions of the Intermediate Value Theorem.
7
(4) Repeat until you reach desired precision (which is determined at each step by looking
at the length of the interval).
Exercise 5.3. Show that cos(x) + x = 2 has a solution in [0, π] and use the bisection method
to estimate the root to within π/4 of its actual value.
Solution. Typically, I like to set things to 0, so we can move the 2 over and consider solutions
to f (x) = cos(x) + x − 2 = 0. This function is continuous on [0, π] since every term is. Next:
f (0) = cos(0) + 0 − 2 = 1 − 2 = −1 < 0,
f (π) = cos(π) + π − 2 = π − 3 > 0,
so f (0) < 0 < f (π). Therefore IVT says there exists a root in [0, π].
To use the bisection method:
(1) The current error is π − 0 = π (length of interval [0, π]). So the midpoint is π/2, and
f (π/2) = cos(π/2) + π/2 − 2 < 0. This tells us we want the subinterval [π/2, π].
(2) The error is now at most π − π/2 = π/2. So at the next step, we will be at our desired
precision. So just take the midpoint and be done: the midpoint is 3π/4, so this is our
approximate root.
Exercise 5.4. Show tan(x) = 1 − x has a solution in (−π/4, π/4), and use the bisection
method to get the root to within ±π/8.
Solution. Again, move everything over (this is my personal preference; you don’t have to):
f (x) = tan(x) + x − 1 = 0. This function is continuous on [−π/4, π/4]. Moreover,
f (π/4) = tan(π/4) +
π
π
− 1 = > 0,
4
4
f (−π/4) = tan(−π/4) −
π
− 1 < 0,
4
so there exists a root in [−π/4, π/4] by IVT. To conclude it must be in the open interval, we
just say that neither endpoint is the root, since we already plugged them in and saw they
didn’t give 0.
Bisection method: Our current error is π/4 + π/4 = π/2. The ±π/8 just means we want an
error of at most π/8 (not double that even though it is ±).
(1) The midpoint is x = 0, and f (0) = tan(0) + 0 − 1 < 0, so we want the interval [0, π/4].
Now our error is at most π/4.
(2) At the next step the error will be π/8, so just take the midpoint to get the approximate
root is π/8.
6
Derivatives
Definition 6.1. Derivative of a function f (x) at a point x = a is
f (a + h) − f (a)
h→0
h
or lim
lim
x→a
8
f (x) − f (a)
,
x−a
provided the limit exists. Derivative as a function:
f 0 (x) = lim
h→0
f (x + h) − f (x)
.
h
Exercise 6.2. Use the limit definition of the derivative to find f 0 (x), where f (x) =
Solution.
f (x + h) − f (x)
h→0
h
√1
− x1
lim x+h
h→0
h
√
√
1
√1
−
x+h x
x
x+h
√
(clear denominators)
lim
√
h→0
h x+h x
√
√
x− x+h
lim √
√
h→0 h( x + h x)
√
√
√
√
( x − x + h)( x + x + h)
√
√
lim
√ √
h→0 h( x + h x)( x +
x + h)
x − (x + h)
√
lim √
√ √
h→0 h( x + h x)( x +
x + h)
−h
√
lim √
√ √
h→0 h( x + h x)( x +
x + h)
−1
√
lim √
√ √
h→0 ( x + h x)( x +
x + h)
−1
√ √ √
(plug in h = 0)
x x(2 x)
−1
√
2x x
−1
.
2x3/2
f 0 (x) = lim
=
=
=
=
=
=
=
=
=
=
Notice this agrees with what we get with the power rule.
Theorem 6.3 (Derivative Rules). Assuming all given functions are differentiable:
(1) [Power Rule] f (x) = xn =⇒ f 0 (x) = nxn−1
(2) [Product Rule] h(x) = f (x)g(x) =⇒ h0 (x) = f 0 (x)g(x) + f (x)g 0 (x)
(3) [Quotient Rule] h(x) =
f (x)
g(x)
=⇒ h0 (x) =
g(x)f 0 (x)−f (x)g 0 (x)
x2
(4) [Chain Rule] h(x) = f (g(x)) =⇒ h0 (x) = f 0 (g(x))g 0 (x)
(5) [Inverses] h(x) = f −1 (x) =⇒ h0 (x) =
1
f 0 (f −1 (x))
9
√1 .
x
(6) [Exponentials] h(x) = ex =⇒ h0 (x) = ex ,
h(x) = ax =⇒ h(x) = ax ln(a) for a > 0
(7) [Logarithms] h(x) = ln(x) =⇒ h0 (x) = x1 ,
h(x) = loga (x) =⇒ h0 (x) =
1
ln(a)
·
1
x
Definition 6.4 (Tangent Line). If y = f (x) is differentiable at x = a, then the tangent line
to x = a is y − f (a) = f 0 (a)(x − a).
Theorem 6.5. If a function f (x) is differentiable at x = a, then f (x) is continuous at
x = a.
Remark. This implies that if f (x) is not continuous at x = a, then f (x) is not differentiable
at x = a.
Exercise 6.6. Find all points a on f (x) = 2x + 3 where the tangent line to f (x) at x = a
goes through (0, 1).
Solution. The tangent line to f (x) at any point is just f (x) since it is a linear function. Since
(0, 1) is not on the line, no tangent line goes through this point.
Exercise 6.7. Find all points a on the graph of f (x) = x2 where the tangent line goes
through (0, −2).
Solution. We can’t be quite as clever as we were in the previous problem. So now we will
build the tangent line to a generic point x = a. Since f 0 (x) = 2x, f 0 (a) = 2a. Also,
f (a) = a2 , so our tangent line is y − a2 = 2a(x − a). To check whether (0, −2) is on the line,
we plug in x = 0, y = −2 into the equation of the line:
√
−2 − a2 = 2a(0 − a) =⇒ −2 − a2 = −2a2 =⇒ a2 = 2 =⇒ a = ± 2.
√
In both cases, f (a) = 2, so there are two points: (± 2, 2).
Exercise 6.8. Find a so that the tangent lines to f (x) = x3 − 2x2 + x + 1 at x = a and
x = a + 1 are parallel.
Solution. Parallel lines have the same slope, so we are looking for the tangent lines to x = a
and x = a + 1 to have the same slope. So we need
f 0 (a) = f 0 (a + 1).
Since f 0 (x) = 3x2 − 4x + 1, we are solving
3a2 − 4a + 1 = 3(a + 1)2 − 4(a + 1) + 1.
Expanding and canceling terms we get
0 = 6a − 1,
so a = 1/6.
Exercise 6.9. The tangent
line to y = f (x) at x = 1 has equation y = 3x + 6. Find the
p
tangent line to g(x) = f (x) at x = 1.
10
Solution. We need both g(1) and g 0 (1): g(1) =
the chain rule, g 0 (1) = √1
2
f (1)
p
f (1) and since g 0 (x) = √1
2
f (x)
0
f 0 (x) by
f 0 (1). The problem is we know neither f (1) nor f (1) at the
moment. To get them, we use the tangent line we are given. Since the tangent line has slope
3, f 0 (1)√= 3. And since the line goes through (1, f (1)), f (1) = 3 · 1 + 6 = 9. Therefore
g(1) = 9 = 3, and
1
1
1
·3= .
g 0 (1) = p
f 0 (1) =
2·3
2
2 f (1)
Therefore the tangent line is y − 3 = 12 (x − 1).
Implicit Differentiation Problems:
2
Exercise 6.10. If y 2 = ex + 2y, find
dy
.
dx
Solution. Since we can’t (or rather don’t want to) solve for y in terms of x, this is implicit
differentiation. Rememeber to use chain rule everywhere:
2
0
2y · y = 2xe
x2
0
0
0
x2
+ 2y =⇒ 2yy − 2y = 2xe
0
=⇒ (2y − 2)y = 2xe
x2
2xex
.
=⇒ y =
2y − 2
0
Exercise 6.11. Find the tangent line to x3 − xy + y 3 = 7 at (2, 1).
Solution. This is again implicit:
3x2 − (xy 0 + y · 1) + 3y 2 y 0 = 0.
Notice we had to use product rule for the middle term. Since all I want is the slope at (2, 1),
I can plug in (2, 1) now:
12 − 2y 0 − 1 + 3y 0 = 0 =⇒ y 0 = −11.
So the tangent line is y − 1 = −11(x − 2).
Exercise 6.12. Find all points on x2 + xy + y 2 = 3 where the tangent line is horizontal.
Solution. Tangent line being horizontal means y 0 = 0. However, we need to (once again) use
implicit differentiation since we can’t/don’t want to solve for y in terms of x:
2x + xy 0 + y + 2yy 0 = 0 =⇒ (x + 2y)y 0 = −2x − y =⇒ y 0 =
−2x − y
.
x + 2y
If y 0 = 0, then −2x − y = 0, so y = −2x. To get the actual point, we plug this back into the
equation for the curve:
x2 + x(−2x) + (−2x)2 = 3 =⇒ 3x2 = 3 =⇒ x = ±1.
Since y = −2x, we get two points: (1, −2) and (−1, 2).
Inverse Problems:
11
Exercise 6.13. If f (x) =
d −1
f (x)
dx
x3
,
4
find
d −1
f (x)
dx
, the tangent line to f −1 (x) at x = 16, and
x=16
using formula (5) in the derivative rules theorem above.
2
Solution. In the formula we need f 0 , so f 0 (x) = 3x4 . Now we don’t want to plug in 16: this
is an input for the inverse so it is an output for the original f . To get the original input, we
solve
x3
f (x) = 16 =⇒
= 16 =⇒ x3 = 64 =⇒ x = 4.
4
So x = 4 is what we plug into f 0 (x) to get
f 0 (4) =
3 · 42
= 12.
4
So
d −1 1
= .
f (x)
dx
12
x=16
For the tangent line, we just found the slope. The point is (16, f −1 (16)) = (16, 4), so the
line is
1
y − 4 = (x − 16).
12
Finally, to find the derivative of the inverse, we know
d −1
1
f (x) = 0 −1
,
dx
f (f (x))
so we need f −1 (x). Switch x and y and solve for y:
x=
So
y3
=⇒ y = (4x)1/3 .
4
1
d −1
f (x) = 0 −1
=
dx
f (f (x))
1
=
3((4x)1/3 )2
4
Exercise 6.14. If f (x) = etan(x) , −π/2 < x < π/2, find
4
.
3(4x)2/3
d −1
f (x)
dx
x=e
√
3
.
Solution. Again, first find f 0 (x): f 0 (x)√= sec2 (x)etan(x) . Now we need to know what to plug
in. So we need the input which gave e 3 as the output:
√
√
e 3 = etan(x) =⇒ 3 = tan(x) =⇒ x = π/3.
√
Since f 0 (π/3) = sec2 (π/3)etan(π/3) = 4e 3 , we get
d −1 1
f (x) √ = √ .
dx
x=e 3
4e 3
Exercise 6.15. If f (x) = tan(x), −π/2 < x < π/2, find
12
d −1
f (x).
dx
Solution. We need f 0 (x) again, so f 0 (x) = sec2 (x). Therefore
1
d −1
f (x) =
.
2
dx
sec (tan−1 (x))
We don’t want to leave it like this, so use the identity 1 + tan2 (x) = sec2 (x):
d −1
1
1
1
f (x) =
=
=
.
−1
2
−1
2
dx
sec (tan (x))
1 + tan (tan (x))
1 + x2
7
Logarithmic Differentiation
We use log differentiation in two main cases: when there are x’s in the base and the exponent
of a function, or when taking logs and simplifying would make the derivative easier, as in
the case of
2√
ex x + 12x
√
y=
.
sec2 (x) sin2 (x) 4 x2 + 1
Exercise 7.1. Find the derivative of y = (2x + 1)sin(x) .
Solution. We need to use log differentiation because we have x’s in the base and the exponent.
So take logs of both sides:
ln(y) = ln((2x + 1)sin(x) ) = sin(x) ln(2x + 1).
Taking derivatives, and remembering implicit on the left:
1 0
2
· y = sin(x) ·
+ ln(2x + 1) cos(x),
y
2x + 1
so
2
+ ln(2x + 1) cos(x)
y = y sin(x) ·
2x + 1
0
sin(x)
= (2x+1)
2
+ ln(2x + 1) cos(x) .
sin(x) ·
2x + 1
Remark. You CANNOT use the ax 7→ ax ln(a) rule here; this only works for a a positive
constant. Make sure you remember this.
Exercise 7.2. If f (x) is a differentiable function, find the derivative of xf (x) .
Solution. Again, write y = xf (x) and take logs:
ln(y) = ln(xf (x) ) = f (x) ln(x).
Taking derivatives:
so
1 0 f (x)
y =
+ ln(x)f 0 (x),
y
x
f (x)
f (x)
0
0
f (x)
0
y =y
+ ln(x)f (x) = x
+ ln(x)f (x) .
x
x
13
8
Linearization
Definition 8.1. The linearization L(x) to f (x) at x = a is L(x) = f (a) + f 0 (a)(x − a).
Remark. Please remember that this is just saying that the y in the tangent line equation is
a good approximation for the function. They are the same thing.
Definition 8.2. We use ∆f for the “absolute error” of f (x). By the
formula:
linearization
∆f ∆f 0
∆f ≈ f (a)∆x. The expression f is the relative error, and 100 f is the percent error
of f .
Exercise 8.3. Find the linearization to f (x) = sin(πx) at x = 1, and use it to approximate
sin(.99π).
Solution. We use the formula: f (1) = sin(π) = 0. Since f 0 (x) = π cos(πx), f 0 (1) =
π cos(π) = −π. Therefore L(x) = 0 − π(x − 1). To approximate sin(.99π), observe this
is what we get if we plug in .99 into f (x), so the approximation is
L(.99) = −π(.99 − 1) = .01π.
Exercise 8.4. The radius of a circle is measured to be 3 with 2% error. Estimate the %
error in calculated area.
. Now,
Solution. The percent error in A = πr2 is 100 ∆A
A
0
∆A A (r)∆r 100 ≈ 100 A A(r) since ∆A ≈ A0 (r)∆r. In this case, r is going to be the 3 which was measured. However, all
we know is that
∆r 100 = 2.
r
So the next step is always multiply by r/r:
0
0
0
A (r)∆r A (r)∆r · r ∆r A (3)3 = 100 100 rA(r) = 100 r A(3) .
A(r) Now we just calculate:
A(3) = 9π,
A0 (r) = 2πr =⇒ A0 (3) = 6π,
so we have an approximation of
3 · 6π = 4,
2 · 9π so the approximate error is 4%.
Exercise 8.5. If we want the error in calculated area of the circle to be no more than 4%,
find the maximum possible % error in radius measurement.
14
Solution. They are the same steps as in the previous solution, but the problem is we are not
given r anymore. So we fix
a radius, which we can say to be our measured radius. I’ll call
∆A this r0 . We want 100 A ≤ 4. Do the same steps:
0
A (r0 )∆r ∆A ≤4
≤ 4 =⇒ 100 100 A A(r0 ) 0
A (r0 )∆r · r ≤4
=⇒ 100 A(r0 ) · r 0
∆r A (r0 )r0 ≤ 4.
=⇒ 100 r
A(r0 ) Now A(r0 ) = πr02 and A0 (r0 ) = 2πr0 , so
0
∆r A (r0 )r0 ≤4
100 r
A(r0 ) =⇒
=⇒
=⇒
∆r 2πr0 r0 ≤4
100 r
πr02 ∆r 100 · 2 ≤ 4
r
∆r 100 ≤ 2.
r
So the maximum allowable percent error is 2%.
9
Extrema and Graphing
Theorem 9.1 (Extreme Value Theorem). If f (x) is continuous on a closed and bounded
interval [a, b], then f (x) attains a global maximum and minimum on the interval.
Theorem 9.2 (Fermat’s Theorem). If f (x) has a local extremum at an interior point c of
[a, b], then f 0 (c) = 0.
Remark. It must be an interior point, this does not apply to endpoints.
Theorem 9.3 (Mean Value Theorem). If f (x) is continuous on [a, b] and differentiable on
(a)
(a, b), then there exists a point c in (a, b) with f 0 (c) = f (b)−f
.
b−a
Theorem 9.4 (Rolle’s Theorem). If f (x) is continuous on [a, b] and differentiable on (a, b)
and f (a) = f (b), then there exists c in (a, b) with f 0 (c) = 0.
Remark. This is MVT with f (a) = f (b).
Method 9.5. To find absolute/global extrema of f (x) on [a, b]:
(1) Find critical points of f (x) (i.e. where f 0 (x) = 0 or is undefined)
(2) Plug all critical points and endpoints into f (x) to determine highest and lowest y-values.
Remark. If the interval is not closed, you still do step (1). But now, plug in whatever
endpoint is included (if any), and take limits to the open endpoints to compare values.
15
Method 9.6 (Local Extrema). To determine whether a critical point x = c is a local extrema,
you can use either the first derivative test (the number line method) or the second derivative
test. The second derivative test says if f 00 (c) > 0, then x = c is a local min (since we’re
concave up), and if f 00 (c) < 0, then x = c is a local min (since we’re concave down). If
f 00 (c) = 0 you cannot conclude anything.
Remark. The second derivative tests only works if f 0 (c) = 0, it does not work where the first
derivative is undefined or at endpoints. Also, if f 00 (c) = 0, do NOT conclude that x = c is
an inflection point; you must check the second derivative number line method to conclude
anything about that.
√
Exercise 9.7. Find all global extrema of f (x) = 3 x on [−1, 8].
Solution. The interval is closed and the function is continuous on the interval, so we can
follow the method outline above for global extrema on the closed interval. First find critical
points: f 0 (x) = 3x12/3 . Notice f 0 (x) is undefined when x = 0, and it is in the interval, so we
care about it. Next we need where f 0 (x) = 0, but this cannot happen since the numerator is
1 6= 0. Therefore the only critical points are x = 0 and the endpoints. Now plug these into
the original function:
f (0) = 0, f (−1) = −1, f (8) = 2.
Therefore the global max is at x = 2 and the global min is at x = −1.
2
Exercise 9.8. Find and classify the global extrema of f (x) = e−x on [−2, 1).
Solution. The interval is not closed, so we will need to do more work. Again, we find critical
2
points: f 0 (x) = −2xe−x . This is defined everywhere, and is 0 only when x = 0. So we have
one critical point. Notice f (0) = e−0 = 1 and f (−2) = e−4 < 1. Finally, we take a limit to
the open endpoint:
2
lim− e−x = e−1 .
x→1
−1
Notice e is between e
global max.
−4
and 1, so x = −2 is going to be the global min and x = 0 the
Exercise 9.9. Find and classify the global extrema of f (x) = x4 − 2x2 + 1 on (−∞, ∞).
Solution. This is sort of the worst case scenario. Again, find critical points: f 0 (x) = 4x3 −
4x = 4x(x2 − 1), so x = 0, ±1 are critical. We have f (±1) = 0 and f (0) = 1. Since
lim f (x) = ∞,
x→±∞
we have no global max, but a global min at x = ±1.
Remark. If you wanted to stick with the book’s methodology, you should show x = ±1 are
local mins, x = 0 is a local max, and then take limits to conclude anything about being
global.
Exercise 9.10. Do the conditions of the MVT apply to the function
sin(πx) if x ≥ 0
f (x) =
πx
if x < 0
on [−1, 1]? If so, find c satisfying the conclusion of the theorem.
16
Solution. First we need continuity on [−1, 1]: The only possible problem is x = 0, so we
check continuity there. Observe that f (0) = sin(π · 0) = 0. Next,
lim πx = 0,
x→0−
lim sin(πx) = sin(π · 0) = 0,
x→0+
so the limit exists, and is equal to f (0). Therefore f (x) is continuous at x = 0, hence on the
whole interval.
Next we need differentiability on the whole open interval. Again, the only possible problem
is x = 0:
π cos(πx) if x > 0
0
f (x) =
.
π
if x < 0
To check differentiability at x = 0, notice
lim π = π,
x→0−
lim π cos(πx) = π,
x→0+
so limx→0 f 0 (x) = π. Therefore f (x) is differentiable at x = 0, and also on (−1, 1). So the
hypotheses are satisfied.
To find c:
sin(π) − (−π)
π
f (b) − f (a)
=
= ,
b−a
1+1
2
0
so we want c with f (c) = π/2. So either π cos(πc) = π/2 or π = π/2. The second one is
clearly not going to help, so we need the first equation to be true. That means cos(πc) = 1/2,
so πc = π/3, or c = 1/3.
Exercise 9.11. Use IVT to show that x5 + 5x − 1 has a real root, and use Rolle to show
that it has exactly one.
Solution. Let f (x) = x5 + 5x − 1. Since f (0) = −1 < 0 and f (1) = 5 > 0, we IVT says there
is a root in [0, 1]. Since f 0 (x) = 5x4 + 5 > 0 is never zero, Rolle’s theorem says there can’t
be more than one root, for if there were two, there would need to be a horizontal tangent
between them.
Method 9.12 (Graphing). To graph a function f (x), follow the steps:
(1) Things to do with f (x):
(a) Make note of any domain restrictions. Remember that you can only plug nonnegative (i.e. 0 or positive) numbers into square roots, and only positive numbers in
logarithms.
(b) Check any asymptotes. They come in three forms:
(i) Vertical: Typically check where denominators are zero, but also remember
that ln(x) has an asymptote at x = 0.
(ii) Horizontal: End behavior, check limits limx→±∞ f (x). If you get a number
for either limit, there is a horizontal asymptote. It does not need to be the
same for both limits.
17
2
+1
(iii) Oblique: This is when you have a rational function (e.g. xx+1
) where the
degree on top is 1 higher than the degree on the bottom. To find it, do long
division and identify the quotient without remainder; it will be a linear term.
(2) Things to do with f 0 (x): Find critical points, determine where f (x) is increasing/decreasing
and any local extrema it may have. As soon as you find critical points and classify them,
find the y-coordinate since you’ll need to graph them.
(3) Things to do with f 00 (x): Find critical points here too, and determine where f (x) is
concave up/down and any inflection points it may have. Again, find the y-coordinate
of such points.
Exercise 9.13. Graph f (x) =
x2 −2
.
x−1
Solution. This has a vertical asymptote at x = 1, and you can check that
lim f (x) = ∞,
lim f (x) = −∞.
x→1−
x→1+
Long division says
1
,
x−1
so we have an oblique asymptote at y = x + 1. Since
f (x) = x + 1 −
f 0 (x) = 1 +
1
,
(x − 1)2
we see that it is undefined at x = 1 but there are no other critical points. Region testing
gives that f (x) is increasing on (−∞, 1) ∪ (1, ∞). Next,
f 00 (x) =
−2
,
(x − 1)3
which is again undefined at x = 1 but never equal to 0. Region testing gives f (x) is concave
up on (−∞, 1) and concave down on (1, ∞).
Exercise 9.14. Graph f (x) = xe−x
10
2 /2
.
L’Hopital’s Rule
There are seven indeterminate forms:
0 ∞
, , 0 · ∞, 00 , 1∞ , ∞0 , ∞ − ∞.
0 ∞
Theorem 10.1. Suppose limx→c
f (x)
g(x)
=
0
0
or
∞
,
∞
where c is allowed to be ±∞. Then
f (x)
f 0 (x)
= lim 0
,
x→c g(x)
x→c g (x)
lim
provided the right hand side exists.
18
So L’Hopital’s rule is simple: provided we have one of those two forms, we just take the
derivative of the top and the bottom (separately; note no quotient rule is needed there) and
then re-evaulate the limit. So for example, since the limit above gives 0/0, we see
(sin(x))0
sin(x)
cos(x)
= lim
= 1.
lim
=
lim
x→0
x→0
x→0
x
(x)0
1
Exercise 10.2. Use L’Hopital’s rule to show:
x2 + 4
=0
x→∞
ex
lim
ex − 1
= 1.
x→0
x
and
lim
In the 0 · ∞ case, we “flip” one of the terms.
Exercise 10.3. Calculate limx→∞ x sin(1/x)
Solution. First observe that it is 0 · ∞. We write x =
1
1
x
to get
sin(1/x)
.
x→∞
1/x
lim x sin(1/x) = lim
x→∞
Now it is 0/0, so we can use L’Hopital:
sin(1/x)
lim
= lim
x→∞
x→∞
1/x
−1
x2
cos(1/x)
= lim cos(1/x) = cos(0) = 1.
x→∞
− x12
Remark. In general, you want to avoid flipping ln unless you’re desperate. Also, flipping
exponentials is generally favorable.
In the exponential cases, we take logs:
x
Exercise 10.4. What is limx→∞ 1 + x1 ?
x
Solution. Let L = limx→∞ 1 + x1 . Taking logs gives
x x
1
1
1
ln(L) = ln lim 1 +
= lim ln 1 +
= lim x ln(1 + ).
x→∞
x→∞
x→∞
x
x
x
Now our new limit is a 0 · ∞ case, so we flip the x:
ln(1 + x1 )
1
lim x ln(1 + ) = lim
.
1
x→∞
x→∞
x
x
Using L’Hopital:
lim
x→∞
ln(1 + x1 )
1
x
= lim
x→∞
1
1+ x1
·
− x12
So out final limit is L = e1 = e.
19
−1
x2
1
x→∞ 1 +
= lim
1
x
= 1.
Finally, in the ∞ − ∞ case, who knows?
√
√
Exercise 10.5. What is limx→∞ x + 1 − x?
Solution. To make a fraction, we can multiply by the conjugate to get:
√
√ √
√
√
√
( x + 1 − x)( x + 1 + x)
1
√
lim x + 1 − x = lim
= lim √
√
√ = 0.
x→∞
x→∞
x→∞
x+1+ x
x+1+ x
11
Antiderivatives
The only way you really know how to do antiderivatives is to guess. The only rule you
1
xn+1 .
“know” is the power rule: xn 7→ n+1
Exercise 11.1. Find the antiderivative of f (x) = sec(5x) tan(5x).
Solution. You know the derivative of sec(x) is sec(x) tan(x), so your initial guess should
be F (x) = sec(5x) + C. The problem is when we take the derivative, we would have 5
coming out from the chain rule which we don’t want, so we cancel that by dividing by it:
F (x) = 15 sec(5x) + C. This is your answer.
The other way such a question could be phrased is by giving a differential equation:
Exercise 11.2. If
dN
dt
= t + sin(1 − t) with N (1) = 52 , find N (t).
Solution. The antiderivative of both sides: The t gives t2 /2. For the sin(1 − t) term, the
“guess” is − cos(1 − t). However, when I take the derivative of this, I get sin(1 − t) · −1 from
the chain rule, so we cancel that by dividing by −1, leaving us with just cos(1 − t). Therefore
2
N (t) = t2 + cos(1 − t) + C. To solve for C, plut in t = 1, and we should get N (1) = 5/2:
5
1
= + cos(0) + C =⇒ C = 1.
2
2
Thus, N (t) =
t2
2
+ cos(1 − t) + 1.
= N + sin(1 −
Remark. Understand the difference between the equation given above and dN
dt
N ). The independent variable t was the only one appearing on the right side, which is
why I could take antiderivatives. If y or N or whatever function you have on the left also
appears on the right, you DO NOT do antiderivatives. It should be that you already know
the solution (the exponential or sin or cos equation you memorized for the second midterm).
Exercise 11.3. Find the general antiderivative of f (x) =
1
.
5−x
Solution. The guess should be ln(5 − x), but when I take a derivative, a negative comes out,
so cancel it out by dividing by −1: F (x) = − ln(5 − x) + C. Now, ln(5 − x) is only defined
for x < 5, but f (x) is defined for all x 6= 5, so we remedy this by putting absolute values:
F (x) = − ln |5 − x| + C. But this also isn’t technically correct. So to be precise, write out
two separate antiderivatives by splitting up the absolute value:
− ln(5 − x) + C
if x < 5
− ln |5 − x| + C =
.
− ln(−(5 − x)) + C if x > 5
20
12
Optimization and Related Rates
I’ll do one of each here:
Exercise 12.1. Find the point on y =
√
x closest to (4, 0).
Solution. (1) Figure out what your optimizing: In this case I want the closest point,
whichpmeans we are minimizing distance to (4, 0). Distance formula says this is
D = (x − 4)2 + y 2 . But minimizing this is the same as minimizing the square of
the distance, so we can instead look at D2 = L = (x − 4)2 + y 2 .
√
(2) Get in terms of one variable: In this case this is fine, since we know y = x: L =
(x − 4)2 + x.
√
(3) Get the interval for your variable: Since x is defined on [0, ∞), this is our interval
for x.
(4) Find the global max or min of the function on this interval: In this case we want the
min. Find critical points:
L0 (x) = 2(x − 4) + 1 = 0 =⇒ x = 7/2.
< 16.
So this is our critical point. Observe that L(0) = 16 and L(7/2) = 41 + 72 = 15
4
Finally, since limx→∞
p L(x) = ∞, we see x = 7/2 is the global minimum. The point on
the graph is (7/2, 7/2).
Exercise 12.2. A particle moves on the circle of radius 2 centered at the origin. At the
moment the particle’s x position is x = 1 and its y-coordinate is negative, dx
= −1. Find
dt
the rate at which the particle’s y-coordinate is changing. Is it increasing or decreasing?
Solution. Draw a picture if you can. In this case the graph is that of x2 + y 2 = 4. Since we
know dx/dt and are looking for dy/dt, this is the relation we will use. Differentiate to get
2x
dx
dy
+ 2y
= 0.
dt
dt
Now we look at the moment we want: x = 1, and to get y we plug into our relation:
√
12 + y 2 = 4 =⇒ y = ± 3,
√
but since y is negative, we have y = − 3. Now plug everything in:
√ dy
dy
1
2 · 1 · −1 + 2(− 3) = 0 =⇒
= −√ .
dt
dt
3
Since it is negative, the y-coordinate is decreasing.
21