Pre-Class Problems 20 Limits Involving Infinity
Discussion on Infinite Limits
Discussion on Limits at Infinity
These are the type of problems that you will be working on in class.
You can go to the solution for each problem by clicking on the problem letter.
1.
Find the following limits.
a.
e.
h.
2.
5
x3
lim
x 3
b.
lim
5
x3
f.
lim
x 3
x 2  5x
lim
16  x 2
x4
c. lim
x 3
5
x 3
x 2  5x
16  x 2
x4
d.
t  10
lim
(t  8) 4
t 8
w9
g. lim
w3
w 2  6w  9
3x 2  12 x
lim
x 2  8 x  16
x4
Find the following limits.
a. lim
x
d.
g.
3x  2
x4
lim
w 
lim
7w  4
2  3w 2
3  4x
x 
9x 2  1
b.
e.
lim
x 
lim
x 
5 x 2  3x  1
3x 2  4 x  8
x3  9
2x 2  x  3
4
h.
lim
x 
c. lim
t 
f.
lim
3t 3  t  2
7  2t 2  6t 3
t  
3
8  t4
27t 4  t 3
3x 12  64 x 8
2  5x 3
SOLUTIONS:
1a.
lim
5
x3
lim
5
5

x3 0
x 3
x 3
Back to Problem 1
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1850
The answer to this one-sided limit is either  or  . We need to find the sign of the function
5
on the immediate right side of 3.
y
x3
Sign of y 
5
:
x 3
X
+

3
Answer: 
1b.
lim
5
x3
lim
5
5

x3 0
x 3
x 3
Back to Problem 1
The answer to this one-sided limit is either  or  . We need to find the sign of the function
5
on the immediate left side of 3.
y
x3
Sign of y 
5
:
x 3

X

3
Answer:  
1c.
lim
5
x 3
lim
5
5

x 3 0
x 3
x3
Back to Problem 1
The answer for this limit is either  ,   , or DNE. You need to calculate the two one-sided
limits. NOTE: An answer of  still tells us the limit does not exist. However, it provides more
information. Namely, that the two one-sided limits go off to the same signed infinity. This is
also true for an answer of  .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1850
From Problem 1a above, we have that lim
x 3
5
  . From Problem 1b above, we have that
x 3
5
   . Since the one-sided limits go off to different signed infinities, then we say
x 3
x 3
5
that lim
= DNE.
x 3 x  3
lim
Answer: DNE
1d.
lim
t 8
lim
t 8
t  10
Back to Problem 1
(t  8) 4
t  10
(t  8)

4
2
0
The answer for this limit is either  ,   , or DNE. You need to calculate the two one-sided
t  10
limits. We need to find the sign of the function y 
on the immediate left and right
(t  8) 4
sides of  8.
Sign of y 
t  10
(t  8) 4
:
X
+

- 10
Thus, lim 
t 8
t  10
(t  8)
4
then we say that lim
t 8
  and lim 
t 8
t  10
(t  8) 4
t  10
(t  8) 4
+

-8
  . Since the signed infinities are the same,

Answer: 
1e.
lim
x4
lim
x4
x 2  5x
Back to Problem 1
16  x 2
x 2  5x
16  x 2

16  20  4

16  16
0
The answer to this one-sided limit is either  or  . We need to find the sign of the function
x 2  5x
y 
on the immediate left side of 4.
16  x 2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1850
Sign of y 
x( x  5)
:
(4  x)(4  x)
X

-4
Thus, lim
x4
x 2  5x

X

0
X

4
X

5

16  x 2
Answer:  
1f.
lim
x 2  5x
16  x 2
lim
x 2  5x
16  20  4


2
16  x
16  16
0
x4
x4
Back to Problem 1
The answer to this one-sided limit is either  or  . We need to find the sign of the function
x 2  5x
y 
on the immediate right side of 4.
16  x 2
Sign of y 
x( x  5)
:
(4  x)(4  x)
X
X

-4
X

0
+

4
X

5
x 2  5x

Thus, lim
x4
16  x 2
Answer: 
1g.
lim
w3
lim
w3
w9
Back to Problem 1
w 2  6w  9
w9
w  6w  9
2

6
6

9  18  9
0
The answer for this limit is either  ,   , or DNE. You need to calculate the two one-sided
w9
limits. We need to find the sign of the function y  2
on the immediate left and
w  6w  9
right sides of 3.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1850
Sign of y 
w 9
( w  3) 2

:


3
Thus, lim
w3
w9
w  6w  9
2
   and lim
w3
are the same, then we say that lim
w3
X

9
w9
w  6w  9
2
w9
w 2  6w  9
   . Since the signed infinities

Answer:  
1h.
lim
x4
lim
x4
lim
x4
3x 2  12 x
Back to Problem 1
x 2  8 x  16
3x 2  12 x
x  8 x  16
3x 2  12 x
2
x  8 x  16
2
=
48  48
0

16  32  16 0
=
lim
x4
3x( x  4)
( x  4)
2
= lim
x4
 12
3x
=
x  4
0
The answer for this limit is either  ,   , or DNE. You need to calculate the two one-sided
3x
limits. We need to find the sign of the function y 
on the immediate left and right
x  4
sides of  4 .
Sign of y 
3x
:
x  4

+

4
Thus, lim 
x4
Thus, lim
x4
3x 2  12 x
x 2  8 x  16
3x 2  12 x
x 2  8 x  16
   and
lim 
x4
X

0
3x 2  12 x
x 2  8 x  16
 .
= DNE
Answer: DNE
2a.
lim
x
3x  2
x4
Back to Problem 2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1850
The largest exponent on the variable x in the numerator and denominator is 1. To change the
3x  2
form of the fraction
in order to get fractions of the form in the theorem above, multiply
x4
the numerator and denominator of this fraction by 1 over x raised to this largest exponent of 1.
1
That is, multiply the numerator and denominator of the fraction by .
x
Thus,
1
2
3
3x  2 x
3x  2
x = 30 3

= lim
= lim
lim
x x  4
x


x x  4
1
4
1 0
1
x
x
Answer: 3
2b.
5 x 2  3x  1
lim
x    3x 2  4 x  8
Back to Problem 2
The largest exponent on the variable x in the numerator and denominator is 2. To change the
5 x 2  3x  1
form of the fraction
in order to get fractions of the form in the theorem above,
3x 2  4 x  8
multiply the numerator and denominator of this fraction by 1 over x raised to this largest
1
exponent of 2. That is, multiply the numerator and denominator of the fraction by 2 . Thus,
x
1
3
1
5  2
5 x 2  3x  1 x 2
5 x 2  3x  1
x x = 500 5

lim
= lim
= lim
2
2
x    3x  4 x  8
x 
x    3x  4 x  8
1
4
8
300 3
3  2
2
x x
x
Answer:
2c.
lim
t 
5
3
3t 3  t  2
7  2t 2  6t 3
Back to Problem 2
The largest exponent on the variable t in the numerator and denominator is 3. In order to change
3t 3  t  2
the form of the fraction
, multiply the numerator and denominator of this fraction
7  2t 2  6t 3
1
by 3 . Thus,
t
1
1
2
3 2  3
3
3t 3  t  2
3t 3  t  2
t
t = 300  3 =  1
lim
 t = lim
= lim
2
3
2
3
t   7  2t
t   7  2t
t  7
1
2
006 6
 6t
2
 6t
 6
3
3
t
t
t
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1850
Answer: 
2d.
lim
w 
1
2
7w  4
2  3w 2
Back to Problem 2
The largest exponent on the variable w in the numerator and denominator is 2. In order to
7w  4
change the form of the fraction
, multiply the numerator and denominator of this
2  3w 2
1
fraction by 2 . Thus,
w
1
7
4
 2
2
7w  4
7w  4
00
0
 w = lim w w =
= lim
lim

0
2
2
w    2  3w
w 
w    2  3w
1
2
03 3
3
w2
w2
Answer: 0
2e.
lim
x 
x3  9
2x 2  x  3
Back to Problem 2
The largest exponent on the variable x in the numerator and denominator is 3. In order to change
x3  9
the form of the fraction
, multiply the numerator and denominator of this fraction
2x 2  x  3
1
by 3 . Thus,
x
1
9
1 3
3
x3  9
x3  9
1 0
1
x
 x = lim
lim
= lim
=

2
2
x   2x  x  3
x  2
x   2x  x  3
1
1
3
0 0  0 0
 2  3
3
x x
x
x
Since you can only approach positive infinity (  ) from one side (the left side), then this limit is
a one-sided limit and the answer to the limit is either  or   . We must determine the sign of
y
x3  9
on the interval containing positive infinity.
2x 2  x  3
Sign of y 
x3  9
:
( x  1) (2 x  3)
X
X

39
X

3

2
Answer: 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1850
+

1
NOTE: Limits approaching negative infinity (   ) are also one-sided limits since you can only
approach negative infinity from the right side.
2f.
lim
t  
3
8  t4
27t 4  t 3
Back to Problem 2
By continuity of the root function, we can pass the limit sign inside the radical sign. Thus, we
have that
lim
t  
3
8  t4
=
27t 4  t 3
3
8  t4
t    27t 4  t 3
lim
The largest exponent on the variable t in the numerator and denominator is 4. In order to change
8  t4
the form of the fraction
, multiply the numerator and denominator of this fraction by
27t 4  t 3
1
under the radical. Thus,
t4
1
8
1
4
4
4
4
4
8t
8t
8t
t
t
3
3
lim
lim
=
= 3 lim
= 3 lim
=

t  
t    27t 4  t 3
t    27t 4  t 3
t  
1
1
27t 4  t 3
27 
t
t4
3
0 1
=
27  0
3
Answer: 
2g.
lim
x 
1
1

27
3
1
3
3  4x
Back to Problem 2
9x 2  1
The largest exponent on the variable x in the numerator is 1. The largest exponent on the
variable x in the denominator is 2, which is under the square root sign. Thus, you must apply the
square root to this exponent by applying the square root to x 2 . Since x is approaching positive
infinity, then
x2  x
 x . Thus, the largest exponent on the variable x in the denominator
is 1. In order to change the form of the fraction
3  4x
9x  1
2
, multiply the numerator and
1
1
. Remember, you will have to square the fraction
in order
x
x
to pass it under the square root sign. Thus,
denominator of this fraction by
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1850
lim
3  4x
x 
9x 2  1
04
90
4

2h.
lim
x 

9
Answer: 
4
= lim
1
3  4x
 x = lim
x 
2
9x  1 1
x
3
4
x
= lim
x 
1
( 9x 2  1)
2
x
4
3
4
3
3x 12  64 x 8
Back to Problem 2
2  5x 3
x 
3
4
x
=
1
9 2
x
The largest exponent on the variable x in the numerator is 12, which is under the fourth root sign.
Thus, you must apply the fourth root to this exponent by applying the fourth root to x 12 . Since x
is approaching negative infinity, then 4 x 12  x 3  ( x) 3   x 3 . Thus, the largest
exponent on the variable x in the numerator is 3. The largest exponent on the variable x in the
4
denominator is 3. In order to change the form of the fraction
numerator and denominator of this fraction by
raise the fraction 
3x 12  64 x 8
, multiply the
2  5x 3
1
1
  3 . Remember, you will have to
3
 x
x
1
to the fourth power in order to pass it under the fourth root sign. Thus,
x3
4
4
lim
3x 12  64 x 8
2  5x 3
x 
4
lim
x 
=
lim
x 
1
( 3x 12  64 x 8 )
12
x
=
2
5 3
x
3x 12  64 x 8
2  5x 3
4
lim
x 
64
x4
=
2
5 3
x
3
4
4
lim
x 
30
50
4
=
3
4
Answer:
4
1
 3
x =

1
 3
x
5
5
x 3 x  3
Example Find lim
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1850
 1 
12
8
  3  ( 3x  64 x )
 x 
=
2
 3 5
x
3
5
Let f ( x) 
5
x3
x
3.1
3.01
3.001
3.0001
f(x)
50
500
5000
50,000
x
2.9
2.99
2.999
2.9999
f(x)
 50
 500
 5000
 50,000
Definition Let f be a function defined on the interval (a, a + r), r > 0. (Right-side of x = a) If
as x  a+, f(x) gets “larger and larger positively”, then we write lim f ( x)   .
x  a
Definition Let f be a function defined on the interval (a, a + r), r > 0. (Right-side of x = a) If
as x  a+, f(x) gets “larger and larger negatively”, then we write lim f ( x)    .
x  a
Definition Let f be a function defined on the interval (a  r, a), r > 0. (Left-side of x = a)
If as x  a, f(x) gets “larger and larger positively”, then we write lim f ( x)   .
x  a
Definition Let f be a function defined on the interval (a  r, a), r > 0. (Left-side of x = a)
If as x  a, f(x) gets “larger and larger negatively”, then we write lim f ( x)    .
x  a
Definition lim f ( x)   if and only if lim f ( x)   and lim f ( x)  
x  a
x  a
xa
Definition lim f ( x)    if and only if lim f ( x)    and lim f ( x)   
x  a
xa
x  a
Thus, for our example above, we may write lim
x 3

5
5
  , lim
   , and
x 3 x  3
x3
5
Does Not Exist (DNE).
x 3 x  3
lim
NOTE: If a one-sided limit of a function has the form
nonzero
, then the answer to the limit
0
will either be  or  . The sign of the infinity will depend on the sign of the function.
Back to the problems
3  x  2x 2
Example Consider the function f ( x) 
for “large” values of x.
x2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1850
x
10
100
1000
10,000
100,000
1,000,000
f(x)
- 1.87
- 1.9897
-1.998997
- 1.99989997
- 1.9999899997
- 1.999998999997
x
f(x)
- 10
- 2.07
- 100
- 2.0097
- 1000
- 2.000997
- 10,000
- 2.00009997
- 100,000
- 2.0000099997
- 1,000,000 - 2.000000999997
Definition Let f be a function defined on the interval (a,  ), where a is a real number. The
statement lim f ( x)  L means as x approaches (positive) infinity (“as x gets larger and larger
x 
positively”), f(x) approaches L.
Definition Let f be a function defined on the interval (  , b), where b is a real number. The
statement lim f ( x)  L means as x approaches negative infinity (“as x gets larger and larger
x
negatively”), f(x) approaches L.
3  x  2x 2
3  x  2x 2
and


2
lim
  2.
x 
x
x2
x2
Thus, for our example above, we may write lim
NOTE:
3  x  2x 2
3
x
2x 2
3
1



 2   2 for all x  0 .
2
2
2
2
x
x
x
x
x
x
c
0
x xr
Theorem If r is a positive rational number and c is any nonzero real number, then lim
c
 0 provided that x r is defined in the later case.
x xr
and lim
Back to the problems
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1850