LESSON 3 THE LIMIT OF A FUNCTION
3x 3  6 x 2  5 x  10
Example Consider the function f ( x ) 
.
x2
The domain of f is the set of all real numbers except for 2. What is the value of f
when x is close to 2?
x
2.5
2.1
2.01
2.001
2.0001
2.00001
2.000001
2.0000001
2.00000001
2.000000001
2.0000000001
2.00000000001
2.000000000001
f(x)
13.75
8.23
7.1203
7.012003
7.00120003
7.0001200003
7.000012000003
7.00000120000003
7.0000001200000003
7.000000012000000003
7.00000000120000000003
7.0000000001200000000003
7.000000000012000000000003
3 ( 2.001) 3  6 ( 2.001) 2  5 ( 2.001)  10
NOTE: f ( 2.001) 
=
2.001  2
3 (8.012006001)  6 ( 4.004001)  10.005  10
=
0.001
24.036018003  24.024006  10.005  10
0.007012003
=
= 7.012003
0.001
0.001
3x 3  6 x 2  5 x  10
It appears that as x is getting closer to 2, f ( x ) 
is getting
x2
closer to 7. We let x get closer to 2 through values that are larger than 2. What
about values of x which are smaller than 2?
Copyrighted by James D. Anderson, The University of Toledo
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x
1.5
1.9
1.99
1.999
1.9999
1.99999
1.999999
1.9999999
1.99999999
1.999999999
1.9999999999
1.99999999999
1.999999999999
f(x)
1.75
5.83
6.8803
6.988003
6.99880003
6.9998800003
6.999988000003
6.99999880000003
6.9999998800000003
6.999999988000000003
6.99999999880000000003
6.9999999998800000000003
6.999999999988000000000003
3 (1.99 ) 3  6 (1.99 ) 2  5 (1.99 )  10
NOTE: f (1.99 ) 
=
1.99  2
3 (7.880599 )  6 ( 3.9601 )  9.95  10
23.641797  23.7606  9.95  10
=
 0.01
 0.01
 0.068803
= 6.8803
 0.01
3x 3  6 x 2  5 x  10
Again, it appears that as x is getting closer to 2, f ( x ) 
x2
getting closer to 7.
is
3x 3  6 x 2  5 x  10
Does the rational expression
simplify? It will simplify if
x2
x  2 is a factor of 3x 3  6 x 2  5x  10 . You might try factoring
3x 3  6 x 2  5x  10 by grouping together the first two terms, which have a 3x 2
in common, and the last two terms, which have a 5 in common.
3x 3  6 x 2  5x  10 = 3x 2 ( x  2 )  5 ( x  2 ) = ( x  2 ) ( 3x 2  5 )
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NOTE: We obtained the final factorization (product) by factoring out the common
x  2 factor in the expression 3x 2 ( x  2 ) and in the expression 5 ( x  2 ) .
3x 3  6 x 2  5 x  10
( x  2 ) ( 3x 2  5 )
Thus,
=
= 3x 2  5 for all x such that
x2
x2
x  2.
3x 3  6 x 2  5 x  10
What does the graph of f ( x ) 
look like? It is the graph of
x2
y  3x 2  5 with the point ( 2 , 7 ) removed.
This graph was created using Maple.
Notation: The mathematical expression x  a means “x approaches a”.
Thus, f ( x )  L means “ f ( x ) approaches L”.
For the example above, we may write the statement: As x  2 , f ( x )  7 . We
are going to define this statement as a limit using the following definitions.
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Informal Definition of a Limit: To say that the limit of f ( x ) is L as x
f ( x )  L , means that f ( x ) gets closer and closer
approaches a , denoted by xlim
a
to L (that is, f ( x ) approaches L) as x gets closer and closer to a (that is, x
approaches a).
f ( x )  L means “As x  a , f ( x )  L .”
That is, xlim
a
3x 3  6 x 2  5 x  10
 7 in the example above, then
Since as x  2 , f ( x ) 
x2
3x 3  6 x 2  5 x  10
 7.
f ( x )  7 or xlim
we may write xlim
2
2
x2
Formal Definition of a Limit: To say that the limit of f ( x ) is L as x approaches
f ( x )  L , means given any   0 , there exists a   0 such
a , denoted by xlim
a
that f ( x )  L   whenever 0  x  a   .
NOTE: That condition that x  a  0 in the definition above means that x  a .
NOTE: The expression
f ( x )  L is the distance between f ( x ) and L on the
real number line and the expression x  a is the distance between x and a on the
real number line. Thus, the condition that f ( x )  L   whenever
0  x  a   is saying that whenever the distance between x and a is less than
 and x  a , then the distance between f ( x ) and L is less than  .
NOTE: The inequality f ( x )  L   is equivalent to the inequality
   f ( x )  L   and this inequality is equivalent to the inequality
L    f ( x )  L   . This last inequality says that f ( x ) is in the interval
(L   , L   ).

L

L

L
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x  a   is equivalent to the inequality
Similarly, the inequality
a    x  a   . This inequality says that x is in the interval ( a   , a   ) .
Thus, the inequality 0  x  a   is equivalent to saying that
a    x  a   and x  a . This says that x is in the set
(a   , a)  (a, a   )

a

a

a
f ( x)  f (a) .
Sometimes, xlim
a
Examples Find the following limits if they exist.
1.
lim ( 3x  5 )
x 1
lim ( 3x  5 )  3  5  8
x 1
( 3x  5 )  8 , this means that as x is getting
Since we are saying that xlim
1
closer to 1 ( x  1 ), 3 x  5 is getting closer to 8 ( 3 x  5  8 ). Let’s see if
this is true. Let f ( x )  3 x  5 .
f (1.1)  3 (1.1)  5  3.3  5  8.3
f (1.01)  3 (1.01)  5  3.03  5  8.03
f (1.001)  3 (1.001)  5  3.003  5  8.003
f (1.0001)  3 (1.0001)  5  3.0003  5  8.0003
f ( 0.9 )  3 ( 0.9 )  5  2.7  5  7.7
f ( 0.99 )  3 ( 0.99 )  5  2.97  5  7.97
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f ( 0.99 9)  3 ( 0.999 )  5  2.997  5  7.997
f ( 0.99 99)  3 ( 0.9999 )  5  2.9997  5  7.9997
( 3x  5 )  8 using the formal definition of limit, we need
To show that xlim
1
to show that for any   0 , there exists a   0 such that 3x  5  8  
whenever 0  x  1   . Since 3x  5  8 = 3x  3 = 3 x  1 ,
then 3x  5  8   whenever 0  3 x  1   , which implies that
3x  5  8   whenever 0  x  1 
choose  

.
3

3
. Thus, given any   0 ,
Answer: 8
2.
lim ( 8 x 2  x  4 )
x2
lim ( 8 x 2  x  4 )  32  2  4  38
x2
( 8 x 2  x  4 )  38 , this means that as x is
Since we are saying that x lim
2
2
getting closer to  2 ( x   2 ), 8x  x  4 is getting closer to 38
2
( 8x  x  4  38 ). Check it out to see if this is true.
( 8 x 2  x  4 )  38 using the formal definition of limit,
To show that x lim
2
we need to show that for any   0 , there exists a   0 such that
8 x 2  x  4  38   whenever 0  x  2   .
2
2
NOTE: 8x  x  4  38 = 8x  x  34 = ( x  2 ) ( 8 x  17 )
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Since we are approaching  2 in a limiting sense, we could restrict ourselves
to the interval described by the inequality x  2  1. This inequality is
equivalent to the inequality  3  x   1 . Thus, if  3  x   1 , then
 24  8 x   8 and  41  8 x  17   25 . Since  41  8 x  17   25 ,
2
then 25  8x  17  41 . Since 8x  x  4  38 = ( x  2 ) ( 8 x  17 )
2
from above, then 8 x  x  4  38 = ( x  2 ) ( 8x  17 ) =
x  2 8x  17 < 41 x  2 . Thus, 8 x 2  x  4  38   whenever
41 x  2   , which implies that
x2 
8 x 2  x  4  38  
whenever

  
. Thus, given any   0 , choose   min  1, 
41
 41
Answer: 38
3.
h 2  4h  3
lim
h  0 6h 2  h  9
h 2  4h  3
003
3
1
lim




h  0 6h 2  h  9
009
9
3
h 2  4h  3
1
lim


Since we are saying that h  0 2
, this means that as h  0 ,
3
6h  h  9
h 2  4h  3
1
 
2
3
6h  h  9
Answer: 
4.
1
3
lim 8
x4
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lim 8  8
x4
NOTE: The limit of a constant is that constant.
Answer: 8
5.
3x 3  6 x 2  5 x  10
lim
x2
x2
3x 3  6 x 2  5 x  10
24  24  10  10
0
lim


x2
x2
22
0
0
is called an indeterminate form because you can not determine
0
any information from it.
NOTE:
We saw in the example at the beginning of this lesson that
3x 3  6 x 2  5x  10 can be factored by grouping and we obtained that
3x 3  6 x 2  5x  10 = ( x  2 ) ( 3x 2  5 ) . Thus, we have that
3x 3  6 x 2  5 x  10
( x  2 ) ( 3x 2  5 )
lim
lim
lim ( 3x 2  5 ) = 7
=
=
x2
x2
x2
x2
x2
Answer: 7
6.
t 2  25
lim
t5 t  5
t 2  25
25  25
0
lim


 Indeterminate Form
t 5 t  5
55
0
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(t  5)(t  5)
t 2  25
lim
lim
( t  5)  10
= t5
= tlim
t5 t  5
5
t 5
Answer: 10
7.
lim
w
lim
w
lim
w
2
3
2
3
2
3
3w  2
4  9w 2
3w  2
22
0


 Indeterminate Form
44
0
4  9w 2
3w  2
3w  2
1
1
1
lim
lim



22
4
4  9w 2 = w  2 ( 2  3w) ( 2  3w ) = w  2 2  3w
3
3
ab
  1 , then you can cancel the 3w  2 and 2  3w as
NOTE: Since
ba
long as you provide a negative one in either the numerator or the
denominator. For this problem, it easier to supply  1 in the numerator.
Answer: 
8.
1
4
x 2  5 x  24
lim
x  3 4 x 2  7 x  15
x 2  5 x  24
9  15  24
0
lim


 Indeterminate Form
x  3 4 x 2  7 x  15
36  21  15
0
We can make use of a theorem from algebra called the Factor Theorem.
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Factor Theorem If p is a polynomial in variable x and p ( r )  0 , then
x  r is a factor of the polynomial p.
2
2
Thus, x  3 is a factor of x  5 x  24 and 4 x  7 x  15 .
2
Thus, x  5 x  24 = ( x  3 ) ( x  8 ) and
4 x 2  7 x  15 = ( x  3 ) ( 4 x  5 )
( x  3) ( x  8)
x8
11
x 2  5 x  24
lim
lim

Thus, xlim
=
=
x  3 ( x  3) ( 4x  5 )
x  3 4x  5
 3 4 x 2  7 x  15
17
Answer:
9.
11
17
3r 2  14r  8
lim
r   4 r 2  4r  32
3r 2  14r  8
48  56  8
0
lim


 Indeterminate Form
r   4 r 2  4r  32
16  16  32
0
By the Factor Theorem given above, r  (  4 )  r  4 is a factor of
3r 2  14r  8 and r 2  4r  32 . Thus,
( r  4 ) ( 3r  2 )
3r  2  10 5
3r 2  14r  8
lim
lim
lim
= r   4 ( r  4 ) ( r  8 ) = r   4 r  8   12  6
r   4 r 2  4r  32
Answer:
5
6
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10.
8t 2  22t  21
lim
3 12t 2  t  6
t
4
9 33 42


8t  22t  21
2
2
2  0 
lim

2
Indeterminate Form
3 12t  t  6
9 3 24
0
t
 
4
4 4
4
2
3
2
is a factor of 8t  22t  21
4
2
and 12t  t  6 . This is equivalent to 4t  3 being a factor of
8t 2  22t  21 and 12t 2  t  6 . Thus,
By the Factor Theorem given above, t 
8t 2  22t  21
( 4t  3 ) ( 2t  7 )
2t  7
lim
lim
lim
=
3 12t 2  t  6
3 ( 4t  3 ) ( 3t  2 ) =
3 3t  2 =
t
t
t
4
4
4
3
7
 6  28
 34
2
=  9  8 =  17  2
9
 2
4

Answer: 2
11.
24 x 2  4 x 3
lim
x  6 x 2  12 x  36
2
3
In order to evaluate 24 x  4 x for x  6 , it is easier to use the factorization
2
3
2
3
2
for 24 x  4 x since 24 x  4 x = 4 x ( 6  x ) . Since the factor 6  x is
2
3
equal to zero when x  6 , then 24 x  4 x = 0 when x  6 . Thus,
24 x 2  4 x 3
0
0
lim 2


 Indeterminate Form
x  6 x  12 x  36
36  72  36
0
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24 x 2  4 x 3
4x 2 ( 6  x )
lim
= xlim
x  6 x 2  12 x  36
6
( x  6 )2
We can cancel the factor of 6  x in the numerator with one on the two
factors of x  6 in the denominator as long as we supply a  1 in either the
numerator or the denominator. Thus, we have that
4x 2 ( 6  x )
4 x 2 (  1)  144
lim
= xlim
=
x6
6
x6
( x  6 )2
0
Now, we need to make use of the following theorem, which will be proved at
the end of this lesson.
f ( x)
lim
lim
f
(
x
)

0
lim
g
(
x
)

0
Theorem If x  a
and x  a
, then x  a g ( x ) does not
exist (DNE).
Using this theorem, we have that
24 x 2  4 x 3
4x 2 ( 6  x )
4 x 2 (  1)  144
lim
 DNE
= xlim
= xlim
=
x  6 x 2  12 x  36
6
6
x6
( x  6 )2
0
Answer: DNE
12.
lim
3w  67  7
w6
lim
3w  67  7

w6
w6
w6
 18  67  7

66
49  7
0

0
0
The only available algebra step to try to simplify the fraction is to rationalize
the numerator of the fraction. Thus, we have that
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lim
w6
3w  67  49
lim
w6
lim
( w  6 ) ( 3w  67  7 )
h2
Answer:
3w  67  7
3w  67  7
=
3w  18
( w  6 ) ( 3w  67  7 )
= wlim
6
( w  6 ) ( 3w  67  7 )
3
14
13.
= wlim
6
3( w  6 )
w6
lim
3w  67  7

w6
3w  67  7
= wlim
6
w6
3
=
3
=
3w  67  7
49  7
=
3
14
5  8h  11
h2
Just like in the previous example, we will need to rationalize the numerator.
Thus,
lim
h2
lim
h2
lim
h2
lim
h2
5  8h  11

h2
5  8h  11
= hlim
2
h2
5  ( 8h  11)
( h  2)( 5 
8h  11 )
= hlim
2
8h  11 )
= hlim
2
16  8h
( h  2)( 5 
8
8
5 
8h  11
=
5 
5
5 
8h  11
5 
8h  11
=
5  8h  11
( h  2)( 5 
8h  11 )
=
8( 2  h )
( h  2)( 5 
= 
8
2 5
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= 
8h  11 )
4
5
=
Answer: 
14.
lim
x3
lim
x3
4
5
7  3x  4
5 x 2  36  9
7  3x  4
5 x 2  36  9
16  4
=
81  9
0
0
=
We need to rationalize both the numerator and the denominator in the
fraction. Thus,
lim
x3
lim
x3
lim
x3
lim
x3
lim
x3
7  3x  4
5 x 2  36  9
7  3x  4
5 x  36  9
2
=

7  3x  4
7  3x  4
5 x 2  36  9

5 x 2  36  9
( 7  3x  16 ) ( 5 x 2  36  9 )
( 5 x 2  36  81) ( 7  3x  4 )
=
(  3x  9 ) ( 5 x 2  36  9 )
( 5 x 2  45 ) ( 7  3x  4 )
=
 3 ( x  3 ) ( 5 x 2  36  9 )
5 ( x 2  9 ) ( 7  3x  4 )
=
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
 3 ( x  3 ) ( 5 x 2  36  9 )
lim
x3
lim
5 ( x  3 ) ( x  3 ) ( 7  3x  4 )
 3 ( 81  9 )
 3 ( 5 x 2  36  9 )
x3
=
 3 (18 )
=
=
5(  6) (8) =
5 (  6 ) ( 16  4 )
5 ( x  3 ) ( 7  3x  4 )
1 ( 18 )
3 (18 )
1( 9 )
9
=
=
=
5 ( 6 ) ( 8 ) 5 ( 2 ) ( 8 ) 5 (1 ) ( 8 ) 40
15.
lim
h0
16 x  16h  25 
h
Answer:
9
40
16 x  25
COMMENT:
We will need to calculate this limit in order to find the
derivative of the function y  16 x  25 using the definition of derivative.
lim
16 x  16h  25 
h
16 x  25
lim
16 x  16h  25 
h
16 x  25
lim
16 x  16h  25 
h
16 x  25
h0
h0
h0
lim
h0
lim
h0
16 x  25 
0
=

0
0
16 x  16h  25 
16 x  25
16 x  16h  25 
16 x  25
16 x  25 )
=
16 x  25 )
=
16h
h ( 16 x  16h  25 
=
=
16 x  16h  25  (16 x  25 )
h ( 16 x  16h  25 
16 x  25
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
=
lim
h0
16
16 x  16h  25 
16
2 16 x  25
16.
16
16 x  25
=
16 x  25 
16 x  25
8
=
16 x  25
=
8
Answer:
16 x  25
( x  h )3  x3
lim
h0
h
COMMENT: We will need to calculate this limit in order to find the
3
derivative of the function y  x using the definition of derivative.
We will use the Binomial Expansion theorem in order to find the four terms
3
in the expansion of ( x  h ) .
Binomial Expansion Theorem Let n be a positive integer. Let a and b be
n
n n  k k
n
n!
n


(
a

b
)

a
b

real numbers, then
, where  k  
k
k !( n  k ) ! .
k 0  
 
Thus, by the Binomial Expansion theorem, we have that
( x  h ) 3  x 3  3x 2 h  3xh2  h 3
( x  h )3  x3
x 3  3x 2 h  3xh2  h 3  x 3
lim
= hlim
=
h0
0
h
h
3x 2 h  3xh2  h 3
h ( 3x 2  3xh  h 2 )
lim
= hlim
=
h0
0
h
h
lim ( 3x 2  3xh  h 2 ) = 3x 2  0  0  3x 2
h0
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
2
Answer: 3x
17.
x 3  64
lim
x4
x4
x 3  64
0
lim

x4
x4
0
3
By the Factor Theorem, x  4 is a factor of x  64 . This factorization can
be found using the factorization formula for the sum of cubes, which is
a 3  b 3  ( a  b ) ( a 2  ab  b 2 )
3
3
3
Since x  64 = x  4 , then replace all the a’s in the formula by x and
replace the b’s in the formula by 4. Thus, we have that
x 3  64 = x 3  43 = ( x  4 ) ( x 2  x4  4 2 ) = ( x  4 ) ( x 2  4 x  16 )
3
The factorization of x  64 can also be found using synthetic division.
3
3
2
Since x  64 = x  0 x  0 x  64 , then we write the following in order
to carry out the synthetic division
1
0
4
0
64
16  64
1
4
16
4
0
3
2
Thus, x  64 = ( x  4 ) ( x  4 x  16 ) . Thus, evaluating the given limit,
we have that
x 3  64
( x  4 ) ( x 2  4 x  16 )
lim
lim
lim ( x 2  4 x  16 ) =
=
=
x4
x4
x4
x4
x4
16  16  16  3 (16 )  48
Answer: 48
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
18.
8t 3  27
lim
3
2t  3
t
2
8t 3  27
0
lim

3
2t  3
0
t
2
3
By the Factor Theorem, 2t  3 is a factor of 8t  27 . This factorization can
be found using the factorization formula for the difference of cubes, which is
a 3  b 3  ( a  b ) ( a 2  ab  b 2 )
3
3
3
Since 8t  27 = ( 2t )  3 , then replace all the a’s in the formula by 2t and
replace the b’s in the formula by 3. Thus, we have that
8t 3  27 = ( 2t ) 3  33 = ( 2t  3 ) ( ( 2t ) 2  ( 2t ) 3  32 ) =
( 2t  3 ) ( 4t 2  6t  9 )
3
The factorization of 8t  27 can also be found using synthetic division.
3
3
2
Since 8t  27 = 8t  0t  0t  27 , then we write the following in order
to carry out the synthetic division
8
0
12
0  27
18
27
8
12
18
3
2
0
3 2

3
t


 ( 8t  12t  18 ) . Factoring out the 2 that is
8
t

27
Thus,
=
2

2
common to all terms in the second factor of 8t  12t  18 , we obtain that
3 2
3


2
8t 3  27 =  t   ( 8t  12t  18 ) =  t   2 ( 4t  6t  9 ) . Now,
2
2


3

distributing the 2 through the first factor of  t   , we obtain that
2

Copyrighted by James D. Anderson, The University of Toledo
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8t 3  27 = ( 2t  3 ) ( 4t 2  6t  9 ) . Thus, evaluating the given limit, we
have that
8t 3  27
( 2t  3 ) ( 4t 2  6t  9 )
lim
lim
lim ( 4t 2  6t  9 ) =
=
=
3
3
3
2t  3
2t  3
t
t
t
2
2
2
9  9  9  3 ( 9 )  27
Answer: 27
f ( x )  L and lim g ( x )  M , and c
Theorem (Properties of Limits) If xlim
a
xa
is any real number, which will called a constant, then
1.
2.
3.
4.
5.
6.
lim c  c
xa
lim c f ( x ) = c lim f ( x )  c L
xa
xa
lim [ f ( x )  g ( x ) ] = lim f ( x ) + lim g ( x )  L  M
xa
xa
xa
lim [ f ( x )  g ( x ) ] = lim f ( x )  lim g ( x )  L  M
xa
xa
xa
lim [ f ( x )  g ( x ) ] =  lim f ( x )   lim g ( x )   L M
 x  a
  x  a

xa
lim f ( x )
L
f ( x)
xa

lim
M  0
=
x  a g( x )
lim g ( x )
M , provided that
xa
7.
lim x n  a n provided that a n is defined
xa
Proof: These properties are proved using the formal definition of limit.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
x 4  6x3  7
Example Find xlim
using properties of limits.
 2 3x 2  x  5
x  6x  7
=
3x 2  x  5
4
lim
x2
3
lim ( x 4  6 x 3  7 )
x2
lim ( 3x  x  5 ) =
2
x2
lim x 4  lim 6 x 3  lim 7
x2
x2
x2
lim 3x  lim x  lim 5 =
2
x2
x2
x2
lim x 4  6 lim x 3  lim 7
16  48  7
24  6 ( 2 )3  7
57
2
=
=
=
2
3 lim x  lim x  lim 5
12  2  5
3( 2 )  2  5
15
x2
x2
x2
x2
x2
x2
f ( x)
f ( x )  0 and lim g ( x )  0 , then xlim
Theorem If xlim
 a g ( x ) does not exist
a
xa
(DNE).
f ( x)
f ( x)
lim
 L . Then
Proof: Suppose that xlim
does
exist,
say
that
 a g( x )
x  a g( x )
 f ( x)


f ( x)  
lim 
 g ( x )  =  lim
g ( x)  = L  0  0.
  xlim

xa
x

a

a
g( x ) 
 g( x )


f ( x)
lim
lim
f
(
x
)

0
This is a contradiction to the fact that x  a
. Thus, x  a g ( x ) does not
exist.
lim f ( x ) =
xa
NOTE: This type of proof in mathematics is called a proof by contradiction. In a
proof by contradiction, an assumption is made and is used along with other
information to obtain a conclusion that contradicts known information. The
assumption, which is made, is usually the opposite of what you want to prove.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850