What Do Limits Have
To Do With Calculus?
An Unlimited Review of Limits
Definition of a Limit
The simplest way to understand a limit of a
function is by looking at it graphically.
Assume we’re looking at this function:
lim(2x  1)
x 2
Let’s see what its graph looks like and what
happens as x approaches 2…
y
3
2
x
It appears as the x-coordinates get closer and closer to 2, the ycoordinates get closer and closer to 3.
We can say that the limit of this function is 3.
A limit is the value a function, f(x), approaches as the variable within that
function gets nearer and nearer to a particular value, x.
Evaluating Limits
Analytically
• Sometimes looking at a graph is not possible, so you
need to have other ways to find a limit.
• We can evaluate limits analytically using algebraic
techniques. These are:
– Substitution
– Factoring (simplifying the expressions)
– Rationalizing the numerator or
denominator (conjugates)
Substitution
• Example 1: Evaluate the following limit:
lim(x 2  4x  9)
x 2
Notice that you can substitute x = 2 into the function
to get 22 – 4 *2 + 9 = 5. Therefore, the limit of this
function as x approaches 2 is 5.
Try this one:
x
lim  cot3x

x 2
2
We can use substitution again -

2  cot3 
2
2
3
cos

2

4 sin 3
2
 0

0
4 1
Now, if all limits were this easy, you would have
worked with them in Algebra I, but…
Factoring
Sometimes substitution just doesn’t work. Let’s look at Example 2:
x 2  3x  4 0
lim

x 1
x 1
0
If you tried using substitution, the result would be 0/0.
The technical term for a result of 0/0 is indeterminate,
which means that you cannot determine the limit using
this method.
What can we do?
Simply do what the title of this slide says: Factor!
(x  4)(x  1)
 lim (x  4)  1  4  5
x 1
x 1
(x  1)
lim
Using Conjugates
The last limit evaluation method
attacks the radical expression in
limits.
This makes applying the conjugate
method rather easy.
Let’s evaluate Example 3:
x 3 2
lim
x 0
x 7
If you tried substitution, it will result in the
indeterminate answer 0/0, and factoring isn’t very
helpful.
To evaluate this limit, multiply the fraction by the
conjugate of the radical expression. But remember
you need to multiply by 1, so you must multiply by the
conjugate/itself.
Let’s look at this:
lim
( x  3  2) 
(x  7)
x 7
 lim
x 7


x 3 2
x 3 2
x 7
(x  7)

x 3 2


  lim
x 7
 lim
x 7

(x  3)  4
(x  7)
1
x 3 2

x 3 2

Now, substitution is possible, and the answer is

1

7-3  2
1
1

4 2 4

Try These
• Do not use your calculator for these.
1.
2.
3.
4.
5.
2x 3  7 x 2  4x
lim
x 4
x 4
x 3
lim
x 9 9  x
x 2  2x  5
lim
x 1
x 1
x3 8
lim
x 2 x  2
x 2  5
,x  4

lim f (x )   5  x
x 4
0, x  4

6. lim
x 0
x
x
Evaluating Limits
Numerically
•
•
By looking at a graph’s table
of data, we can determine the
limit of the function.
Let’s look at Example 4:
X
2.9
2.99
2.999
3.001
3.01
3.1
g(x)
-4.41
-4.9401
-4.994
5.006
5.0601
5.61
From the above data, estimate the value of the following limits:
lim g (x )
x 3
lim g (x )
x 3
lim g (x )
x 3
X
2.9
2.99
2.999
3.001
3.01
3.1
g(x)
-4.41
-4.9401
-4.994
5.006
5.0601
5.61
lim g (x )
x 3
It appears that as we approach 3 from the left, the g(x) values are getting
closer and closer to -5. Therefore, we can assume that lim g (x )  5.
x 3
lim g (x )
x 3
It appears that as we approach 3 from the right, the g(x) values are getting
closer and closer to 5. Therefore, we can assume that lim g (x )  5.
x 3
lim g (x )
x 3
If the limit from the left, lim g (x ), is not equal to the limit from the right,
x 3
lim g (x ), we would have to say that the limit Does Not Exist lim g (x )  D.N.E.
x 3
x 3
Properties of Limits
If c, k, R, S, U, and V are finite numbers and if
limf (x )  R , lim g (x )  S , lim f (x )  U , lim g (x )  V , then
x c
x c
x 
x 
1a. lim kf (x )  kR
1b. lim kf (x )  kU
2a. lim f (x )  g (x )   R  S
x c
2b. lim f (x )  g (x )   U V
x 
3a. lim f (x ) g (x )  RS
3b. lim f (x ) g (x )  UV
x c
x 
x c
x 
f (x )
R

(if S  0)
x c g (x )
S
f (x ) U

(if V  0)
x  g (x )
V
4a. lim
4b. lim
5a. lim k  k
5b. lim k  k
x c
6.
x 
The Squeeze or Sandwich theorem.
If f(x)  g(x)  h (x ) and if lim f (x )  lim h (x )  L, then lim g (x )  L.
x c
x c
x c
Special Limits
Special Trig limits:
7a. lim
sin 
 0
8a.
9.
lim

1
sin a 
a

cos   1
lim
0
 0

 0
A Special e limit:
10.
ex 1
lim
1
x 0
x
7b. lim
 
sin 

0
sin a  a

 0 sin b
b
8b. lim
Rational Function Theorem
We can determine the limit of a quotient of polynomials
(rational functions)
Using the fact that…
i.
1
lim n  0
x  x
P(x)
0
x  Q(x)
When the degree of P(x) is less than that of Q(x), then lim
and y=0 is a horizontal asymptote of the graph.
P(x)
  or -
x  Q(x)
ii. When the degree of P(x) is higher than that of Q(x), then lim
and the graph has no horizontal asymptotes.
P(x) an

x  Q(x)
bn
iii. When the degree of P(x) and Q(x) are the same, then lim
where an and bn are the coefficents of the highest powers of x in P(x) and Q(x)
a
and y = n is a horizontal asymptote of the graph.
bn
Try These
•
Do not use your calculator
1.
2.
3.
4.
5.
6.
lim x 2  5x  11
x 8
x 3
2
x 5 x  15
lim  2
lim
x 0
x 3
2
x 5 x  15
10x 2  25x  1
lim
x 
x4 8
x4 8
lim
2
x  10x  25x  1
lim
5x 4  2x
7. lim
x 
x2
x 2
8.
lim 2
x 6 x  4x  12
x 2
9.
lim 2
x 6 x  4x  12
x 2
10. lim 2
x 6 x  4x  12
11.
12.
13.
14.
15.
x
x
lim
x 0
x
x 7 x  49
x
lim 2
x 7 x  49
x
lim
2
x 7 (x  7)
lim
2
2
x  5, x  3
Let f(x)= 
x  2, x  3
find a.
lim f (x ),
lim f (x ),
lim f (x )
x2  5, x  3
16.
Let f(x)= 
x  1, x  3
find a. lim f (x ), lim f (x ),
lim f (x )
x 3
x 3
17.
18.
19.
20.
lim 3cos x
x

4
x
x  0 cos x
x
lim 3
x  0 sin x
sin3x
lim
x  0 sin 8x
lim 3
x 3
x 3
x 3
x 3
21.
22.
23.
24.
25.
26.
27.
28.
tan 7x
x  0 sin 5x
lim sin x
lim
x 
1
x 
x
x 2 sin x
lim
2
x 0 1  cos x
sin2 7x
lim
2
x 0 sin 11x
(3  h )2  9
lim
h 0
h
sin(x  h )  sin x
lim
h 0
h
1
1

lim x  h x
h 0
h
lim sin
Answers will be posted later.