Limits
By
Dr. Julia Arnold
Definition of Limit
The function f has the limit L as x approaches a,
written
x
lima f(x)  L
Provided the value of f(x) can be made as close
to the number L as we please by taking x
sufficiently close to (but not equal to) a.
Example 1: Let f(x) = x3
Find the lim f ( x)
x
Solution:
2
3
3
lim
x

2
8
x
2
8
y
2
x
Example 2:
x  2.....if...x  1
g(x)  
 1.....if...x  1
Find
x
lim1 g ( x)
y
3
2
1
-4
-3
-2
-1
1
-1
-2
-3
2
3
4
x
5
Example 2:
x  2.....if...x  1
g(x)  
 1.....if...x  1
x
lim1 g ( x)  3
y
3
2
1
-4
-3
-2
-1
1
-1
-2
-3
2
3
4
x
5
 1...if...x  0
Example 3: f(x)  
 1...if...x  0
Find:
x
lim0 f(x)
y
3
2
1
-4
-3
-2
-1
1
-1
-2
-3
2
3
4
x
5
 1...if...x  0
Example 3: f(x)  
 1...if...x  0
Find:
x
lim0 f(x)
Because you get different results
as you come from the right side of
zero
and then from the left side of
y
03 there is no limit.
2
1
-4
-3
-2
-1
1
-1
-2
-3
2
3
4
x
5
Example 4:
1
g( x )  2
x
Find:
x
lim0 g(x)
y
3
2
1
-4
-3
-2
-1
1
-1
-2
-3
2
3
4
x
5
Example 4:
1
g( x )  2
x
Find:
x
lim0 g(x)
If the limit is
infinity from
either side ,
then there is
no limit.
y
3
2
1
-4
-3
-2
-1
1
-1
-2
-3
2
3
4
x
5
Properties of Limits
Suppose
lima f(x)  L
x
Then
1)
And
x
lima g ( x)  M
r


lim
f
(
x
)

lim
f
(
x
)

L
x
a
x
a
2)
3)
r
x
x
limacf ( x)  c x lima f ( x)  cL
lima [ f ( x)  g ( x)] x lima f ( x) x lima g ( x)  L  M
4)
x
5)
r
lima [ f ( x) g ( x)]x lima f ( x) x lima g ( x)  LM
f ( x) x lima f ( x) L
lima

 ,M  0
x
g ( x) x lima g ( x) M
Properties of Limits
Suppose
x
lima x  2
Then
1) lim
x
3
3
x

2
8
2
3
2
3
2
lim4 5x  5[ x lim4 x]  54  58  40
3
2
2)
x
3)
4
4
lim
5
x

2

lim
5
x
 x lim1 2 
x
1
x
1


 
5 x lim1 x  x lim1 2  51  2  3
4
4)
4
3
2
3
2
lim
2
x
x

7

[
lim
2
x
][
lim
x
 7] 
x
3
x
3
x
3


2( x lim3 x)3 x lim3 x 2  7  23
3


9  7  544  216
5)
2 x  1 x lim2 (2 x  1) 24  1 9
lim2


 3
x
2  1 3
x 1
lim2 ( x  1)
x
2
2
Indeterminate Forms
If you are trying to find a limit as x approaches a and
when you substitute a for x in the expression you get
0
then you have an indeterminate form and it is
0
now your job to find a way to determine the limit.

2
4
x
4
Example 1: Find lim
x
2
x 2

This is an indeterminate form since substitution of 2
into the expression gives 0
0
Indeterminate Forms

2
4
x
4
Example 1: Find lim
x
2
x 2

This is an indeterminate form since substitution of 2
into the expression gives 0
0
One way to determine the limit is by factoring the
expression and simplifying


4 x2  4
4x  2x  2
lim

lim
x lim2 4(x  2)  16
x
2
x
2
x 2
x 2
We can see why this is true by graphing the function:
19y
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
x
-1 1 2 3 4 5 6 7 8 91011
-18
-17
-16
-15
-14
-13
-12
-11
-10
-9-8-7-6
- 5-4-3-2-1
1 21314151617
1 81920
-2
-3
-4
-5
-6
-7
-8
-9
Example 2:
h
lim0
1  h 1
h
Substitution of 0 for h in the above expression yields
0/0 and thus an indeterminate form.
For radicals,
try
rationalizing
the
numerator in
this case.
 1  h  1  1  h  1 


lim0 
h




h
1

h

1



 1  h 1 

lim0 
h

h
1

h

1




h


lim0 
h

h
1

h

1



 1
1


lim0 
h
 2
1

h

1








Again let’s view the graph of this function:
In this problem h can be changed to x.
x
lim0
1  x 1 1

x
2
y
3
lim0
h
2
1  h 1 1

h
2
1
-4
-3
-2
-1
1
-1
-2
-3
2
3
4
x
5
Limits at Infinity
Example 1:
2x2
f(x) 
1  x2
1
Since the fraction
x
Find
2x2
lim 
x
1  x2
gets smaller and smaller as
x gets larger and larger, we can conclude that
the limit of 1/x as x approaches  is 0.
We now multiply the fraction above by
1
x2
1
x2
Limits at Infinity
Example 1:
2x2
f(x) 
1  x2
Find
2x2
lim 
x
1  x2
We now multiply the fraction above by
x2
2 2
2
2
x  lim
lim

2
2 x
x


1
1 x
0 1

1

x2
x2 x2
1
x2
1
x2
Let’s look at the graph of this function:
Recall that y = 2 would be a horizontal asymptote
14y
13
12
11
10
9
8
7
6
5
4
3
2
1
x
-1 1 2 3 4 5 6 7 8 91011
-18
-17
-16
-15
-14
-13
-12
-11
-10
-9-8-7-6
- 5-4-3-2-1
1 21314151617
1 81920
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-12
-13
-14
2
x
x 3
Example 2: lim
x

2x3  1
Always choose the highest exponent in the denominator
to divide by: ie. x3
 1
 x2  x  3  x3

lim  
x
3
 2x  1  1
 x3






 x2 x
3 
 3 3 3
x
x 
x
lim
x

 2x3 1 
 3 

3
x 
 x
3 
1 1



2
3 
0
x
x
x


lim 
 0
x
1
 2
 2


x3


2
3
x
 8x  4
Example 3: lim
x

2x2  4x  5
Always choose the highest exponent in the denominator
to divide by: ie. x2
3x2  8x  4
lim 2

x
2x  4x  5
8 4
3  2
x x 3
lim
x

4 5
2  2 2
x x
3
2
2
x

3
x
1
Example 4: lim
x

x2  2x  4
Always choose the highest exponent in the denominator
to divide by: ie. x2
2x3  3x2  1
lim 2

x
x  2x  4
1
2x  3  2
2x  3
x
lim


x
2 4
1
1  2
x x
3
2
2
x

3
x
1
Example 5: lim
x

x2  2x  4
Always choose the highest exponent in the denominator
to divide by: ie. x2
2x3  3x2  1
lim 2

x
x  2x  4
1
2x  3  2
2x  3
x
lim

 
x
2 4
1
1  2
x x
These two examples do not imply that if x  
that the answer is   or if x   that the answer
is  
Example 6: Consider the function f(x) = -x3
y
3
2
3
lim

x
 
x

-4
-3
-2
1
-1
1
-1
-2
-3
2
3
4
x
5
In summary
To find a limit substitute the limiting value
into the function.
Be cautious if the function is defined in a
piece-wise manner.
If you substitute and get 0/0 that’s an
indeterminate form and must be
determined. 0/0 is never an acceptable
answer. Try to factor if the problem
consists of polynomials. Rationalize if the
problem contains radicals.
To find limits at infinity, always divide
every term top and bottom by the highest
degreed term in the denominator.
Go to the homework.