LESSON 11 L’HOPITAL’S RULE AND THE SANDWICH THEOREM
Theorem (Cauchy’s Formula) If the functions f and g are continuous on the
closed interval [ a , b ] and differentiable on the open interval ( a , b ) and if
g ( x )  0 for all x in ( a , b ) , then there is a number c in ( a , b ) such that
f (b)  f (a )
f ( c )

.
g( b )  g( a )
g ( c )
Proof Note that g ( b )  g ( a )  0 . For if g ( b )  g ( a )  0 , then g ( a )  g ( b ) .
Thus, by Rolle’s Theorem, there exists a number w in the open interval ( a , b ) such
that g ( w )  0 . This would contradict the fact that g ( x )  0 for all x in ( a , b ) .
Define the function h on the closed interval [ a , b ] by
h( x )  [ f ( b )  f ( a ) ] g ( x )  [ g ( b )  g ( a ) ] f ( x ) .
Note that f ( b )  f ( a ) and g ( b )  g ( a ) are fixed numbers. Since the functions f
and g are continuous on the closed interval [ a , b ] , then the function h is
continuous on [ a , b ] . Since the functions f and g are differentiable on the open
interval ( a , b ) , then the function h is differentiable on ( a , b ) .
Note that h( a )  [ f ( b )  f ( a ) ] g ( a )  [ g ( b )  g ( a ) ] f ( a ) =
f ( b ) g ( a )  f ( a ) g ( a )  g ( b ) f ( a )  g ( a ) f ( a ) = f ( b ) g ( a )  g ( b ) f ( a ) and
h( b )  [ f ( b )  f ( a ) ] g ( b )  [ g ( b )  g ( a ) ] f ( b ) =
f ( b ) g( b )  f ( a ) g( b )  g( b ) f ( b )  g( a ) f ( b ) = g( a ) f ( b )  f ( a ) g( b )
Thus, by Rolle’s Theorem, there exists a number c in the open interval ( a , b ) such
that h( c )  0 . Since h( x )  [ f ( b )  f ( a ) ] g ( x )  [ g ( b )  g ( a ) ] f ( x ) and
f ( b )  f ( a ) and g ( b )  g ( a ) are constants, then
h( x )  [ f ( b )  f ( a ) ] g ( x )  [ g ( b )  g ( a ) ] f ( x ) .
Thus, h( c )  [ f ( b )  f ( a ) ] g ( c )  [ g ( b )  g ( a ) ] f ( c ) and h( c )  0 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
[ f ( b )  f ( a ) ] g ( c )  [ g ( b )  g ( a ) ] f ( c )  0 
[ f ( b )  f ( a ) ] g ( c )  [ g ( b )  g ( a ) ] f ( c ) 
f (b)  f (a )
f ( c )

g( b )  g( a )
g ( c )
NOTE: Cauchy’s Formula is a generalization of the Mean Value Theorem. If
g ( x )  x , then g ( a )  a , g ( b )  b , and g ( x )  1 for all x. Thus,
f (b)  f (a )
f ( c )
f (b)  f (a )
f ( c )



 f ( b )  f ( a )  f ( c ) ( b  a ) .
g( b )  g( a )
g ( c )
ba
1
Theorem (L’Hopital’s Rule) Suppose the functions f and g are differentiable on
f ( x)
a deleted neighborhood of the number a. If g ( x )  0 for all x  a and if
g( x )
0

has the indeterminate form
or
, then
0

lim
x  a
f ( x)
f ( x )
 lim
.
g ( x ) x  a g ( x )
f ( x )
f ( x )
lim
  .
provided that xlim
exists
or
 a g ( x )
x  a g ( x )
Proof Let { x : 0  x  a   } = ( a   , a )  ( a , a   ) be a deleted
neighborhood of the number a on which the functions f and g are differentiable.
f ( x)
0
Suppose that
has the indeterminate form
at x  a . Then we have that
g( x )
0
f ( x )
 L for some number
lim f ( x )  0 and lim g ( x )  0 . Suppose that xlim
 a g ( x )
x  a
x  a
f ( x)
 L . Let’s defined the following two
L. We want to show that xlim
 a g( x )
functions F and G on the open interval ( a   , a   ) .
 f ( x) , x  a
F( x )  
 0 , x a
 g( x ) , x  a
G
(
x
)


and
 0 , x a
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Since F ( x )  f ( x ) for all x  a in ( a   , a   ) , then the function F is
differentiable on the deleted neighborhood ( a   , a )  ( a , a   ) . Thus, the
function F is continuous on the deleted neighborhood ( a   , a )  ( a , a   ) .
F ( x )  lim f ( x )  0 = F ( a ) , then the function F is continuous at
Since xlim
 a
x  a
x  a . Thus, the function F is continuous on the interval ( a   , a   ) .
Similarly, the function G is continuous on ( a   , a   ) . Thus, for each x in the
deleted neighborhood ( a   , a )  ( a , a   ) , we can apply Cauchy’s Formula to
the interval [ x , a ] if x  a or to the interval [ a , x ] if x  a for the functions F
and G. Thus, by Cauchy’s Formula, there exists a number c x between a and x
such that
F ( x )  F ( a ) F ( c x )

G ( x )  G ( a ) G ( c x )
Since x  a , then we have that F ( x )  f ( x ) and G ( x )  g ( x ) . Since c x  a ,
then we have that F ( c x )  f ( c x ) and G( c x )  g ( c x ) . Also, F ( a )  0 and
G( a )  0 . Thus,
F ( x )  F ( a ) F ( c x )


G ( x )  G ( a ) G ( c x )
f ( c x )
f ( x)  0


g( x )  0
g ( c x )
f ( c x )
f ( x)

g( x )
g ( c x ) .
Since c x is between a and x, then as x  a , we have that c x  a . Thus,
lim
x  a
f ( c x )
f ( c x )
f ( x)
 lim
 lim
 L.
g ( x ) x  a g ( c x ) c x  a g ( c x )
This argument would also be true if xlim
 a
indeterminate form
f ( x)
   . The argument for the
g( x )

is more difficult and can be found in texts on advanced

calculus.
COMMENT: One common mistake, which is made by students applying
L’Hopital’s Rule, is using the Quotient Rule instead of differentiating the
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
numerator and denominator. The other mistake is applying L’Hopital’s Rule
0

without verifying that you have the indeterminate form
or
.
0

Examples Find the following limits, if they exist.
1.
sin x
 0
x
lim
x
sin x
sin 0
0
 Indeterminate Form
=
=
 0
x
0
0
lim
x
Applying L’Hopital’s Rule, we have that
sin x
cos x
lim
cos x = 1
= x0
= xlim
 0
 0
x
1
lim
x
Answer: 1
NOTE: The calculation of this limit was needed in order to calculate the
derivative of the sine function.
2.
t
t
sin 3 t
 0
5t
lim
sin 3 t
sin 0
0
 Indeterminate Form
=
=
 0
5t
0
0
lim
Applying L’Hopital’s Rule, we have that
t
sin 3 t
3 cos 3 t
3
3
lim
lim cos 3 t =
=
=
 0
t  0
5t
5
5 t0
5
lim
Answer:
3
5
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
3.
lim 
x  0
Since
lim 
x  0
sin 2 x
5x
5x 
5
x , then
lim 
x  0
sin 2 x
=
5x
1
5
lim 
x  0
sin 2 x
.
x
sin 2 x
0
 Indeterminate Form
=
0
x
Applying L’Hopital’s Rule, we have that
lim 
sin 2 x
=
5x
1
5
lim  4
x  0
x  0
1
5
lim 
x  0
x cos 2 x =
sin 2 x
=
x
4
5
lim 
x  0
1
5
lim 
x  0
2 cos 2 x
=
1
2 x
x cos 2 x =
4
(0) = 0
5
Answer: 0
4.
lim
cos h  1
h
lim
cos h  1
cos 0  1
11
0
 Indeterminate Form
=
=
=
h
0
0
0
h  0
h  0
Applying L’Hopital’s Rule, we have that
lim
h  0
cos h  1
 sin h
sin h   1 ( 0 )  0
= hlim
=  hlim
 0
 0
h
1
Answer: 0
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
NOTE: The calculation of this limit was also needed in order to calculate the
derivative of the sine function.
5.
1  cos 6 x
 0
4x2
lim
x
1  cos 6 x
1  cos 0
11
0
 Indeterminate Form
=
=
=
2
 0
4x
0
0
0
lim
x
Applying L’Hopital’s Rule, we have that
 (  6 sin 6 x )
1  cos 6 x
6 sin 6 x
3
sin 6 x
lim
lim
lim
=
=
=
2
x  0
 0
x  0
8x
4x
8x
4 x0 x
lim
x
sin 6 x
0
 Indeterminate Form
=
x
0
lim
x  0
Applying L’Hopital’s Rule again, we have that
3
sin 6 x
3
3
9
lim
lim 6 cos 6 x = ( 6 ) =
=
4 x0 x
4 x0
4
2
Answer:
6.
9
2
lim 
1  cos t
t3
lim 
1  cos t
1  cos 0
11
0
 Indeterminate Form
=
=
=
3
t
0
0
0
t  0
t  0
Applying L’Hopital’s Rule, we have that
lim
t  0

1  cos t
=
t3
lim
t  0

 (  sin t )
1
sin t
lim  2
=
2
3t
3 t0 t
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
lim
t  0
sin t
0
 Indeterminate Form
=
2
0
t
Applying L’Hopital’s Rule again, we have that
1
cos t
1
sin t
1
cos t
lim 
lim  2 =
lim 
=
3 t  0 2t
3 t0 t
6 t0
t
lim 
t  0
cos t
1
 This is NOT an indeterminate form.
=
0
t
cos t
:
t
Sign of y 




2

0
cos t
1
sin t
1
cos t
   . Since
lim  2 =
lim 
by
t  0
t
3 t0 t
6 t0
t
1
sin t
lim  2    .
L’Hopital’s Rule, then
3 t0 t
Thus,
lim 
Thus,
lim
t  0

1  cos t
1
sin t
lim  2    .
=
3
t
3 t0 t
Answer:  
7.
lim
tan 4 x
x
lim
tan 4 x
tan 0
0
 Indeterminate Form
=
=
x
0
0
x  0
x  0
Applying L’Hopital’s Rule, we have that
tan 4 x
lim 
=
x  0
x
4 sec 2 4 x
lim 
4 sec 2 0 = 4 ( 1 ) 4
=
x  0
1
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Answer: 4
8.
lim
tan (  6 x )
x  0
3
x5
Since the tangent function is an odd function, then we may write tan (  6 x ) =
 tan 6 x . Thus, we have that
lim
tan (  6 x )
x  0
lim
3
x5
tan 6 x
x  0 3
x
5
=
= xlim
 0
 tan 6 x
3
x5
=  xlim
 0
tan 6 x
3
x5
tan 0
0
 Indeterminate Form
=
0
0
Applying L’Hopital’s Rule, we have that
 lim
tan 6 x
x  0 3
x5
6 sec 2 6 x
18
sec 2 6 x

lim
lim
= x  0 5 2/3 = 
x  0
5
x 2/3
x
3
1
sec 2 6 x
 This is NOT an indeterminate form.
lim
=
x  0
0
x 2/3
sec 2 6 x
Sign of y 
:
x 2/3
+
+

0
sec 2 6 x
  and
Thus, x lim
 0
x 2/3
sec 2 6 x
sec 2 6 x
lim
  . Thus, lim
 .
x  0
x  0
x 2/3
x 2/3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Since  xlim
 0
 lim
tan 6 x
x  0 3
x5
Thus, xlim
 0
18
sec 2 6 x
lim
= 
by L’Hopital’s Rule, then
5 x  0 x 2/3
tan 6 x
3
x5
  .
tan (  6 x )
3
x5
=  xlim
 0
tan 6 x
3
x5
  .
Answer:  
9.
sec 5  1
 0
7 2
lim

sec 5  1
sec 0  1
11
0
 Indeterminate Form
=
=
=
2
 0
7
0
0
0
lim

Applying L’Hopital’s Rule, we have that lim
 0
sec 5  1
=
7 2
5 sec 5 tan 5
 0
14
lim

Now, using Properties of Limits, we have that lim
 0
5 sec 5 tan 5
=
14
5
tan 5 
5
sec 5 tan 5

( lim sec 5 )  lim
lim

=
14   0
14   0

  0  
sec 5  1 , then
Since lim
 0
5
tan 5 

( lim sec 5 )  lim
 =
14   0
  0  
5
tan 5 
5
tan 5

(1)  lim
lim

=
14
14   0 
  0  
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
lim
  0
tan 5
0
 Indeterminate Form
=

0
Applying L’Hopital’s Rule again, we have that
5
tan 5
5
25
25
25
lim
lim 5 sec 2 5 =
lim sec 2 5 =
sec 2 0 =
=
14   0 
14   0
14   0
14
14
Thus, lim
 0
Answer:
10.
sec 5  1
5 sec 5 tan 5
5
tan 5
25
lim
= lim
=
=
.
2
 0
7
14
14   0 
14
25
14
4 x 2  3x  6
lim
x   5  2 x  3x 2

4 x 2  3x  6
 Indeterminate Form
lim
=
x   5  2 x  3x 2

Applying L’Hopital’s Rule, we have that
8x  3

4 x 2  3x  6
lim
 Indeterminate Form
lim
=
=
x   2  6x
x   5  2 x  3x 2

Applying L’Hopital’s Rule again, we have that
lim
x  
8x  3
8
4
lim
= x
= 
2  6x
6
3
Answer: 
4
3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
11.
lim
6t  7
t2  9
lim
6t  7

 Indeterminate Form
=
2
t 9

t  
t  
Applying L’Hopital’s Rule, we have that
lim
t  
6t  7
=
t2  9
lim
t  
6
1
= 3 t lim  = 3 ( 0 )  0
2t
t
Answer: 0
12.
5x  3x 2
lim
x    4 x  11

5x  3x 2
 Indeterminate Form
lim
=
x    4 x  11

Applying L’Hopital’s Rule, we have that
5x  3x 2
lim
=
x    4 x  11
lim
x  
5  6x
1
lim ( 5  6 x ) = 
=
4
4 x  
Answer: 
13.
lim
12  8 w 6  3 w 10
( 4 w 2  5) 3 ( 2 w 4  3)
lim
12  8 w 6  3 w 10

 Indeterminate Form
=
2
3
4
( 4 w  5) ( 2 w  3)

w  
w  
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
You can apply L’Hopital’s Rule to this limit. However, it much easier to use
the methods of the MATH-1850 course to evaluate this limit.
12  8 w  3 w
=
( 4 w 2  5) 3 ( 2 w 4  3)
lim
12
8

3
w 10 w 4
=
1
2
3 1
4
(
4
w

5
)
(
2
w

3
)
w6
w4
lim
12
8

3
w 10 w 4
3
3  =
 1
 
2
(
4
w

5
)
2

 2
 

w4 
w
 
w  
w  
w  
1
10
12  8 w  3 w
w

1
( 4 w 2  5) 3 ( 2 w 4  3)
w 10
6
lim
6
10
lim
w  
10
=
lim
12
8
 4 3
10
w
w
3
3  =
 1 
2
3 
 2  ( 4 w  5)  2  4 
w 
w 

lim
12
8

3
w 10 w 4
3
5  
3  =

4

2


 

w2  
w4 

w  
w  
003
3
3
= 
= 
3
( 4  0) ( 2  0)
64 ( 2 )
128
Answer: 
3
128
NOTE: I will let you do the TEN applications of L’Hopital’s Rule in order to
produce this answer.
14.
lim
5 x  11  3
x 2
lim
5 x  11  3
=
x 2
x  4
x  4
9 3
0
 Indeterminate Form
=
0
4 2
Applying L’Hopital’s Rule, we have that
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
1
( 5 x  11)  1 / 2 5
1/ 2
5 x  11  3
5
x
5 (2)
2
lim
= xlim
=
=
=
1
/
2
 4
1  1/ 2
x  4 ( 5 x  11 )
3
x 2
x
2
lim
x  4
10
3
Answer:
10
3
NOTE: To calculate this limit, without using L’Hopital’s Rule, would
require the algebra of rationalizing both the numerator and the denominator.
15.
3
t 2  49  2 3 3
t 5
3
t 2  49  2 3 3
=
t 5
lim
t  5
lim
t  5
3
 24  2 3 3
23 3  23 3
=
=
0
0
0
 Indeterminate Form
0
Applying L’Hopital’s Rule, we have that
3
lim
t  5
t 2  49  2 3 3
=
t 5
2
t
1 2
lim
( t  49 )  2 / 3 2 t =
=
2
5 3
3 t   5 ( t  49 ) 2 / 3
lim
t 
10 ( 2 ) 3 3
53 3
2
5
10
10 ( 24 ) 1 / 3

= 
= 
= 
= 
=
3 (  24 ) 2 / 3
3 ( 24 )
3 (6)
3 ( 24 ) 2 / 3
3 ( 24 )
53 3

18
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Answer: 
53 3
18
NOTE: This is the limit which is required to calculate the value of the
2
derivative of the function f ( t )  3 t  49 at t   5 using the definition of
derivative. It is ironic that we used the value of the derivative of this function
at t   5 in order to calculate this limit. To calculate this limit, without
using L’Hopital’s Rule, would require the algebra of rationalizing the
3
3
2
3
numerator. Since a  b  ( a  b ) ( a  a b  b ) , then in order to
rationalize the numerator of
3
3
t 2  49  2 3 3 in the fraction
t 2  49  2 3 3
, we would need to multiply the numerator and denominator
t 5
2
2
2
2
of the fraction by ( 3 t  49 )  ( 3 t  49 ) ( 2 3 3 )  ( 2 3 3 ) =
3
( t 2  49 ) 2  2 3 3 ( t 2  49 )  4 3 9 . Thus, we would have that
3
lim
t  5
3
lim
t  5
lim
t  5
lim
t  5
lim
t  5
t 2  49  2 3 3
=
t 5
t 2  49  2 3 3

t 5
3
( t 2  49 ) 2  2 3 3 ( t 2  49 )  4 3 9
3
( t 2  49 ) 2  2 3 3 ( t 2  49 )  4 3 9
( 3 t 2  49 ) 3  ( 2 3 3 ) 3
( t  5 ) ( 3 ( t 2  49 ) 2  2 3 3 ( t 2  49 )  4 3 9 )
=
t 2  49  24
( t  5 ) ( 3 ( t 2  49 ) 2  2 3 3 ( t 2  49 )  4 3 9 )
=
t 2  25
( t  5 ) ( 3 ( t 2  49 ) 2  2 3 3 ( t 2  49 )  4 3 9 )
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
=
=
lim
t  5
( t  5) ( t  5)
( t  5 ) ( 3 ( t 2  49 ) 2  2 3 3 ( t 2  49 )  4 3 9 )
t 5
lim
t  5 3
( t 2  49 ) 2  2 3 3 ( t 2  49 )  4 3 9
=
 10
3
=
 10
(  24 ) 2  2 3 3 (  24 )  4 3 9
=
3
24 2  2 3 3 ( 24 )  4 3 9
=
 10
 10
=
=
( 3 24 ) 2  2 3 3 ( 8  3 )  4 3 9
(2 3 3 ) 2  4 3 9  4 3 9
53 3
53 3
10
 10
5
= 
=  3
=  3
= 
=
6 ( 3)
43 9  43 9  43 9
6 9
6 27
12 3 9
53 3

18
16.
4x  9 
lim
x  3
3
3
21
5 x 2  19  4
4x  9 
lim
x  3
L’Hopital’s Rule was definitely easier.
21
5 x 2  19  4
=
21  21
0
 Indeterminate Form
=
3 64  4
0
Applying L’Hopital’s Rule, we have that
4x  9 
lim
x  3
lim
x  3
3
21
1
( 4 x  9 )  1/ 2 4
2
= xlim
=
 3 1
5 x  19  4
( 5 x 2  19 )  2 / 3 10 x
3
2
2 ( 4 x  9 )  1/ 2
10
x ( 5 x 2  19 )  2 / 3
3
6 ( 4 x  9 )  1/ 2
3
( 5 x 2  19 ) 2 / 3
= xlim
= xlim
=
 3 10 x ( 5 x 2  19 )  2 / 3
5  3 x ( 4 x  9 ) 1/ 2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
16 21
3 64 2 / 3
1 16
16


=
=
=
5 3 21
105
5
21
5 21
Answer:
17.
16 21
105
sin 3 y  cos 5 y  e 3 y
lim
y  0
y2
sin 3 y  cos 5 y  e 3 y
011
0
lim
 Indeterminate Form
=
=
y  0
0
0
y2
Applying L’Hopital’s Rule, we have that
3 cos 3 y  5 sin 5 y  3 e 3 y
sin 3 y  cos 5 y  e 3 y
lim
= ylim
=
 0
y  0
2y
y2
303
0
 Indeterminate Form
=
0
0
Applying L’Hopital’s Rule again, we have that
3 cos 3 y  5 sin 5 y  3 e 3 y
3 (  3 sin 3 y )  5  5 cos 5 y  3  3 e 3 y
lim
= ylim
=
y  0
 0
2y
2
0  25  9
 34
 9 sin 3 y  25 cos 5 y  9 e 3 y
lim
=
=
=  17
y  0
2
2
2
Answer:  17
18.
ln x 3
lim
x  0  cot 6 x
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
ln x 3

lim 
 Indeterminate Form
=
x  0 cot 6 x

3 ln x
ln x 3
lim
Since x lim
=
, then applying L’Hopital’s Rule to the
x  0  cot 6 x
 0  cot 6 x
last limit, we have that
3
3 ln x
1
sin 2 6 x
x
lim
= x lim
=  x lim
2
x  0  cot 6 x
 0   6 csc 6 x
2  0
x
0
sin 2 6 x
 Indeterminate Form
lim 
=
x  0
0
x
Applying L’Hopital’s Rule again, we have that
1
1
sin 2 6 x

lim 2 sin 6 x ( cos 6 x ) 6 =  3 lim  sin 12 x =

lim 
=
x  0
2 x  0
2 x0
x
 3 (0)  0
NOTE: By the double angle formula for sine, which is sin 2  =
2 sin  cos  , then 2 sin 6 x cos 6 x  sin 12 x .
Answer: 0
19.
sin 2 4 t
lim
t  0
3t 2
sin 2 4 t
0
lim
 Indeterminate Form
=
t  0
3t 2
0
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
sin 2 4 t
1
sin 2 4 t
lim
Since tlim
=
, then applying L’Hopital’s Rule to the
 0
3t 2
3 t  0 t2
last limit, we have that
1
2 sin 4 t ( cos 4t ) 4
2
2 sin 4 t cos 4t
1
sin 2 4 t
lim
lim
lim
=
=
=
3 t0
2t
3 t0
t
3 t  0 t2
2
sin 8 t
lim
3 t0 t
NOTE: By the double angle formula for sine, 2 sin 4 t cos 4 t  sin 8 t .
t
sin 8 t
0
 Indeterminate Form
=
 0
t
0
lim
Applying L’Hopital’s Rule again, we have that
2
sin 8 t
2
16
16
16
16
lim
lim 8 cos 8 t =
lim cos 8 t =
cos 0 =
(1) =
=
3 t0 t
3 t0
3 t0
3
3
3
sin 2 4 t
1
2 sin 4 t ( cos 4t ) 4
2
sin 8 t
lim
lim
Thus, tlim
=
=
=
2
 0
3t
3 t0
2t
3 t0 t
16
16
lim cos 8 t =
3 t0
3
Answer:
20.
lim
  
lim
  
16
3

tan 3

ln sin
2

tan 3
tan 3
tan 
0
 Indeterminate Form
=
=
=


ln
1
0
ln sin
ln sin
2
2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Applying L’Hopital’s Rule, we have that
tan 3
lim 
 =
  
ln sin
2
Since
lim  tan
  
3 sec 2 3
lim
 =
   1
cos
2
2

sin
2

   and
2
1 , then 6 lim  tan
  
lim
  

3 sec 2 3

6
lim
tan
sec 2 3
=
1


  
2
cot
2
2
lim  sec 2 3 = sec 2 3 = sec 2  = (  1) 2 =
  

sec 2 3    .
2
Thus, by L’Hopital’s Rule, we have that
lim
  

tan 3
  .

ln sin
2
Answer:  
21.
y3  8
lim
y   2 ln ( 4 y 2  15 )
0
y3  8
0
lim
 Indeterminate Form
=
=
2
y   2 ln ( 4 y
ln 1
 15 )
0
Applying L’Hopital’s Rule, we have that
2
3y2
 15
2 4y
lim
lim
3
y

= y  2
=
y  2
8y
8y
4 y 2  15
3
3
3
lim y ( 4 y 2  15 ) = (  2 ) ( 1 ) = 
8 y  2
8
4
y3  8
lim
=
y   2 ln ( 4 y 2  15 )
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
3
4
Answer: 
22.
lim
ln ( 7  2 x )
x2  6x  9
lim
ln ( 7  2 x )
ln 1
0
 Indeterminate Form
=
=
2
x  6x  9
9  18  9
0
x  3
x  3
Applying L’Hopital’s Rule, we have that
2
2
ln ( 7  2 x )
1
7  2x
2x  7
lim 2
lim
lim
lim
=
=
=
x  3 x
x  3 ( x  3) ( 2 x  7 )
x 3 2x  6
x  3 2 ( x  3)
 6x  9
Sign of y 
1
:
( x  3) ( 2 x  7 )

+

3
NOTE: The domain of the function y 
lim
x  3
Thus, xlim
 3
ln ( 7  2 x )
   and
x2  6x  9
lim
x  3
7
2
ln ( 7  2 x )
is the set of numbers
x2  6x  9
 7
given by (   , 3 )   3,  .
 2
Thus,

ln ( 7  2 x )
 .
x2  6x  9
ln ( 7  2 x )
= DNE.
x2  6x  9
Answer: DNE
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Now, we will study the indeterminate forms of 0   , 1  , 0 0 ,  0 , and    .
Using algebra, we may write the indeterminate form 0   as either the
0

indeterminate form
or the indeterminate form .
0

Using logarithmic functions, we may write the indeterminate forms 1  , 0 0 , and  0
as the indeterminate form 0   .
Examples Find the following limits, if they exist.
1.
lim x 2 e 5 x
x  
Since
lim e 5 x  0 , then
x  
lim x 2 e 5 x =   0  Indeterminate Form
x  
e 5x
x 2 e 5x =
Since x e =  2 , then we could write x lim



x
0
indeterminate form of this last limit is .
0
2
2
Since x e
5x
5x
=
x2
2
e  5x
x e
, then we could write x lim
 
5x

indeterminate form of this last limit is
.

Applying L’Hopital’s Rule to
lim
x  
e 5x
=
x2
lim
x  
lim
x  
=
lim
x  
lim
x  
e 5x
. The
x2
x2
e  5x
. The
e 5x
, we have that
x2
5e 5 x
5
5e 5 x
=  x lim
 2x 3
2   x3
0
5e 5 x
Since x lim
=
, then we can apply L’Hopital’s Rule again.

3
  x
0
However, since the exponent on the x in the denominator will always stay
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
negative and the numerator will always contain the exponential function, we
will not be able to determine an answer for the limit.
Let’s apply L’Hopital’s Rule to
form
lim
x  
lim
x  
lim
x  

. Thus, we have that

x2
e  5x
x
e  5x
lim
=
x  
x2
, which has the indeterminate
e  5x
2x
2
x
lim
 5x = 

 5e
5 x    e 5x

 Indeterminate Form

=
Apply L’Hopital’s Rule again, we have that

2
1
2
x
2
1
2

lim
lim
lim
lim e 5 x =
=
=
=

5
x

5
x

5
x
5 x     5e
5 x   e
25 x    e
25 x   
2
(0)  0
25
Thus,
2
lim x e
5x
lim
=
x  
x2
x  
e  5x
= 
2
x
2
lim
lim e 5 x  0
=

5
x
x



x
5
e
25   
Answer: 0
2.
lim
t  (   / 2) 
Since
sec t ln tan
lim
t  (   / 2) 
5t
2
sec t    and
ln  1 = 0, then
lim
t  (   / 2) 
lim
t  (   / 2)
sec t ln tan
5t
2

ln tan
5t
 5 
 ln tan  
 =
2
 4 
=    0  Indeterminate Form
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
ln tan
Since sec t ln tan
lim
t  (   / 2) 
form of
5t
2
sec t ln tan
=
5t
2
5t
2
5t
2
, then we can write
cos t
ln tan
=
1
sec t
5t
2
, which has an indeterminate
cos t
ln tan
=
lim
t  (   / 2) 
0
.
0
Applying L’Hopital’s Rule, we have that
5t
2
cos t
ln tan
lim
t  (   / 2) 
=
lim
t  (   / 2) 
5
5t
sec 2
2
2
5t
tan
5
2
= 2
 sin t
5t
2
5t =
sin t tan
2
sec 2
lim
t  (   / 2) 
 5 
sec 2  

2
5
4 
(

2
)

5
5
 



(2) =  5
=
=
2
  
 5 
2
(

1
)
(

1
)
2
sin    tan  

 2
 4 
Answer:  5
3.
lim  sin 3 x ln x 2
x  0
Since
lim  ln x 2    and
x  0
lim sin 3 x  0 , then
x  0
lim  sin 3 x ln x 2 =
x  0
0     Indeterminate Form
2
Since sin 3 x ln x
ln x 2
2 ln x
=
=
, then we can write
1
csc 3 x
sin 3 x
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
lim  sin 3 x ln x 2 =
x  0
lim
x  0
2 ln x

, which has an indeterminate form of
.
csc 3 x

Applying L’Hopital’s Rule, we have that
2
2
1
2 ln x
x

lim 
lim 
lim 
=
=
=
x  0 csc 3 x
x  0  3 csc 3 x cot 3 x
3 x  0 x csc 3 x cot 3 x

2
sin 3 x  
2
sin 3 x tan 3 x

  lim 
lim 
  lim  tan 3 x 
=
3x0
x x 0
3 x0
x

lim 
x  0
sin 3 x
0
 Indeterminate Form
=
x
0
Applying L’Hopital’s Rule again, we have that
lim 
x  0
sin 3 x
=
x
lim 3 cos 3x  3
x  0
tan 3 x  tan 0  0 , then 
Since x lim
 0

2
sin 3 x
 lim 
3x0
x


  lim  tan 3 x  =

x 0
2
( 3) ( 0 )  0
3
Answer: 0
4.
lim  cot 
  0
Since
lim cot    , then
  0
lim  cot  = 0    Indeterminate Form
  0
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860


 cot  =
Since  cot  = 1
=
, then we can write  lim
 0
tan 
cot 

0
lim
,
which
has
an
indeterminate
form
of
.
  0 tan 
0


Applying L’Hopital’s Rule, we have that
lim
  0


=
tan 
lim
  0

1
=
sec 2 
lim  cos 2   1
  0
Answer: 1
5.
 6 
lim x 3 sin  3 
x  
x 
6 
 6 
 6 

sin  3  = sin  lim 3  = sin 0  0 , then lim x 3 sin  3 
Since xlim
x  
 
x 
x 
x x 
=   0  Indeterminate Form
 6 
sin  3 
x 
 6 
 6 
3
3
x
sin
lim
x
sin
 3 =


Since
=
,
then
we
can
write
3
1
x  
x 
x 
3
x
 6 
sin  3 
x 
0
lim
,
which
has
an
indeterminate
form
of
.
x  
1
0
x3
Applying L’Hopital’s Rule, we have that
 6 
 6 
 6 
 1 
 6 
cos  3   D x  3 
cos  3   6 D x  3 
sin  3 
x 
x 
x 
x 
x 
lim
lim
lim
= x
= x
=
x  
1
 1 
 1 
D
D




x
x
3
3
x3
x 
x 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
6 
 6 

6 lim cos  3  = 6 cos  lim 3  = 6 cos 0  6 (1)  6
x  
x 
x x 
Answer: 6
6.
lim
y  (   / 2) 
Since
( sec 3 y  tan 3 y )
lim
y  (   / 2) 
lim
y  (   / 2) 
sec 3 y    and
lim
y  (   / 2) 
tan 3 y    , then
( sec 3 y  tan 3 y ) =    (   ) =     Indeterminate
Form
lim
( sec 3 y  tan 3 y ) =
lim
1  sin 3 y
cos 3 y
y  (   / 2) 
y  (   / 2) 
Since
then
lim
y  (   / 2)
lim
y  (   / 2) 

y  (   / 2) 
y  (   / 2) 
 1
sin 3 y


 cos 3 y cos 3 y

 =

 3 
 3 
sin 3 y  sin  
lim  cos 3 y  cos  
  1 and
  0,
y  (   / 2)
 2 
 2 
1  sin 3 y
0
 Indeterminate Form
=
cos 3 y
0
Applying L’Hopital’s Rule to
lim
lim
1  sin 3 y
=
cos 3 y
lim
lim
y  (   / 2) 
y  (   / 2) 
1  sin 3 y
, we have that
cos 3 y
 3 cos 3 y
=
 3 sin 3 y
lim
y  (   / 2) 
Answer: 0
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
cot 3 y  0
7.
lim [ ln ( 2 x  3)  ln (5 x  4 ) ]
x  
ln ( 2 x  3)   and lim ln (5 x  4 )   , then
Since xlim
 
x  
lim [ ln ( 2 x  3)  ln (5 x  4 ) ] =     Indeterminate Form
x  
ln
lim [ ln ( 2 x  3)  ln (5 x  4 ) ] = xlim
 
x  

2x  3 
2x  3
 =
= ln  xlim
  5x  4 
5x  4


2
2

ln  lim  = ln
5
x5
2x  3

has an indeterminate form of
, then we can
5x  4

apply L’Hopital’s Rule.
NOTE: Since xlim
 
2
5
Answer: ln
8.
lim ( 6 x 2  7 x  9 
x  
Since
lim
x  
6 x 2  11 )
6 x 2  7 x  9   and
lim
x  
6 x 2  11   , then
lim ( 6 x 2  7 x  9 
6 x 2  11 ) =     Indeterminate Form
lim ( 6 x 2  7 x  9 
6 x 2  11 ) =
lim ( 6 x  7 x  9 
6 x  11 ) 
x  
x  
2
x  
lim
x  
2
6 x 2  7 x  9  ( 6 x 2  11)
6x2  7x  9 
6 x 2  11
=
6x2  7x  9 
6 x 2  11
6x2  7x  9 
6 x 2  11
lim
x  
=
 7 x  20
6x2  7x  9 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
6 x 2  11
=
1
 7 x  20
 x =
2
2
6 x  7 x  9  6 x  11  1
x

lim
x  
7
lim
x  
1
(6 x 2  7 x  9) 
2
x
7
lim
x  
7
9.
20
x
7
9
6  2 
x x
=
7 6
12
Answer:
7 6
12
2 6
20
x
11
6 2
x
=
1
2
( 6 x  11)
x2
=
70
600 
60
=
7
6 
lim ( t 2  ln t 2 )
t  
Since
t
lim t 2   and
 
lim ln t 2   , then
t  
t
lim ( t 2  ln t 2 ) =
 
    Indeterminate Form
lim ( t  ln t ) =
 
2
t
Now, consider
t

ln t 2
lim t  1  2
 
t

2
2
t




ln t 2
 Indeterminate Form
lim
=
  t2

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
6
=
ln t 2
Since t lim  2 =
t
limit, we have that
lim
t  
Thus,
Thus,
Since
Thus,
2 ln t
=
t2
t
t
t
lim
t  
2
lim t =
   2t
ln t 2
lim
=
  t2
lim
lim t   and
t  
t
t  
1
0
t2
2 ln t
 0.
t2

 = 1  0  1 .

2
t

ln t 2

lim 1  2
  
t

lim ( t  ln t ) =
 
2
lim
t  

ln t 2
lim  1  2
 
t

2 ln t
, then applying L’Hopital’s Rule to this last
t2

ln t 2

lim t  1  2
 
t

2
2
t

  1 , then


ln t 2

lim t  1  2
 
t

2
t

  

Answer: 
10.
1

lim    log 3 (  y ) 
y  0
y

Since
lim
y  0
1
   and
y
lim log 3 (  y )    , then
y  0
1

lim    log 3 (  y )  =     Indeterminate Form
y  0
y

1

lim    log 3 (  y )  =
y  0
y

lim
y  0
1
[1  y log 3 (  y ) ]
y
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

   .

Now, consider
Since
lim y log 3 (  y ) = 0     Indeterminate Form
y  0
lim y log 3 (  y ) =
y  0
lim
y  0
log 3 (  y )
, then applying L’Hopital’s Rule
1
y
to this last limit, we have that
lim
y  0
Thus,
Since
lim
y  0

Thus,
log 3 (  y )
=
1
y
lim
y  0
1
 y ln 3
1
1

(0)  0

lim
y
=
=
1
ln 3
ln 3 y  0 
 2
y
lim y log 3 (  y )  0 . Thus,
y  0
lim
y  0
1
   and
y
lim [1  y log 3 (  y ) ] = 1  0  1 .
y  0
lim [1  y log 3 (  y ) ] = 1, then
y  0
1
[1  y log 3 (  y ) ]    .
y
1

lim    log 3 (  y )  =
y  0
y

lim
y  0

1
[1  y log 3 (  y ) ]   
y
Answer:  
f ( x ) g ( x ) if the indeterminate form is 1  , 0 0 , and
Guidelines for evaluating xlim
 a
0.
1.
2.
g( x )
Let y  f ( x )
Take the natural logarithm of both sides of the equation: ln y  g ( x ) ln f ( x )
Of course, you can use any base for the logarithm. The natural logarithm is
the easiest to use.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
3.
ln y if it exists.
Find xlim
 a
4.
ln y  L , then lim f ( x ) g ( x ) = lim y = e L .
If xlim
x  a
 a
x  a
x
NOTE: Since the natural exponential function y  e is continuous for all
ln y
y =
real numbers and e  y for all positive real numbers, then xlim
 a
lim ln y
lim e ln y = e x  a
x  a
ln y ) = exp ( L ) = e L .
= exp ( xlim
 a
Examples Find the following limits, if they exist.
1.
lim ( 9 x 2  1) 1 / x
2
x  0
( 9 x 2  1)  1 and lim
Since xlim
x  0
 0
1
2
1/ x 2


lim
(
9
x

1
)
, then x  0
=
x2
1   Indeterminate Form.
Let y  ( 9 x  1)
2
1/ x 2
1
ln ( 9 x 2  1)
2
. Then ln y  2 ln ( 9 x  1) =
.
x
x2
0
ln ( 9 x 2  1)
Now, find xlim
,
which
has
an
indeterminate
form
of
.
 0
0
x2
Applying L’Hopital’s Rule, we have that
18 x
9
18 x
1
ln ( 9 x  1)
9x 2  1
lim
9
lim

lim
lim
=
=
=
2
2
2
x  0 9x
x  0 9x
x  0
x  0
1
 1 2x
2x
x
2
ln ( 9 x 2  1)
9
ln y = lim
Thus, xlim
x  0
 0
x2
y = lim ( 9 x 2  1) 1 / x = e 9
Thus, xlim
 0
x  0
2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Answer: e 9
2.
lim  (1  tan 5 t ) csc 3 t
t  0
(1  tan 5 t )  1 and
Since t lim
 0
lim csc 3t    , then
t  0
lim  (1  tan 5 t ) csc 3 t = 1    Indeterminate Form.
t  0
csc 3 t
Let y  (1  tan 5t )
. Then ln y  csc 3 t ln (1  tan 5 t ) =
Now, find t lim
 0
ln (1  tan 5 t )
.
sin 3 t
ln (1  tan 5 t )
0
, which has an indeterminate form of .
sin 3 t
0
Applying L’Hopital’s Rule, we have that
 5 sec 5 t
ln (1  tan 5 t )
5
sec 5 t
1  tan 5 t
lim 

lim
lim
= t  0
=
=
t  0
sin 3 t
3 t  0  (1  tan 5 t ) cos 3 t
3 cos 3t

5
1
5

= 
3 (1  0 ) (1)
3
ln y = lim 
Thus, t lim
t  0
 0
ln (1  tan 5 t )
5
= 
sin 3 t
3
y = lim  (1  tan 5 t ) csc 3 t = e  5 / 3
Thus, t lim

t  0
 0
 5/3
Answer: e
3.
1

lim  1  
x  
x

x
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
1

lim  1  
x  
x

x

= 1  Indeterminate Form.
1

ln  1  
x

1
1


ln
y

x
ln
1

y

1





Let
. Then
=
.
1
x
x


x
x
1

ln  1  
x

0
lim
Now, find x  
, which has an indeterminate form of .
1
0
x
Applying L’Hopital’s Rule, we have that
1

Dx  1  
x

1
1
1


Dx  
ln  1  
1  
x
x
x


lim
lim
lim
= x
= x
x  
1
1 1 =
1
1

Dx  

 Dx  
x
x x

x
x
1
1
1
=
 
1
1

0
1
x
lim
1

ln y = lim x ln  1    1
Thus, xlim
x  
 
x

1

y = lim  1  
Thus, xlim
x  
 
x

x
= e
Answer: e
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
x
x
1
1


1   = e , then we can use the expression  1  

NOTE: Since xlim
 
x
x


to generate approximations for the number e . You can use your calculator to
1



verify that  1 
1
,
000
,
000


4.
t
t
4

lim  1  
 
t 

2t
4

lim  1  
 
t 

2t
1, 000, 000
 2.71828 .

 Indeterminate Form.
= 1
4

2
ln
1



2t
t 

4
4


Let y   1   . Then ln y  2 t ln  1   =
.
1
t 
t 


t
Now, find
t
4

2 ln  1  
t 

0
lim
, which has an indeterminate form of .
 
1
0
t
Applying L’Hopital’s Rule, we have that
t
4

2 ln  1  
t 

lim
= 2 t lim 
 
1
t
4

Dt  1  
t 

4
 4

Dt   
1  
t 
 t 

2
lim
= t   
4  1 =
1
1

Dt  

 Dt  
t  t

t
1
 4 Dt  
1
t
1
2 lim

8
lim

8

= 8
t   
t  
4 =
4  1 =
1

0
1
 1   Dt  
t
t  t

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
4

ln y = lim 2 t ln  1     8
Thus, xlim
x  
 
t 

y =
Thus, xlim
 
t
Answer: e  8 or
5.
4

lim  1  
 
t 

2t
= e8
1
e8
lim  x x
x  0
lim  x x = 0 0  Indeterminate Form.
x  0
Let y  x . Then ln y  x ln x =
x
Now, find
lim
x  0
ln x
1 .
x
ln x

,
which
has
an
indeterminate
form
of
.
1

x
Applying L’Hopital’s Rule, we have that
1
ln x
lim 
lim  x =  lim 1  x 2 =  lim x  0
=
1
1
x  0
x  0
x  0 x
x  0
 2
x
x
ln y =
Thus, xlim
 
y =
Thus, xlim
 
lim x ln x  0
x  0
lim  x x = e 0  1
x  0
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Answer: 1
6.
3t
lim ( e )
5
tan  
 t 
t  
5
Since t lim  e  0 and t lim  tan   = tan 0  0 , then
t
0
= 0  Indeterminate Form.
3t
Let y  ( e )
3t
5
tan  
 t 
3t
lim ( e )
5
tan  
 t 
t  
5
5
3t
ln
y

tan
ln
e
3
t
tan


 .
. Then
=
t
t
5
lim 3 t tan   , which has an indeterminate form of    0 .
t  
t
5
tan  
t
5
lim 3 t tan   = 3 t lim 
, which has an indeterminate form
1
 
t
t
Now, find
Since
of
t
0
, then we can applying L’Hopital’s Rule to this last limit. Thus,
0
5 5
5
1
5
sec 2   D t  
sec 2   5 D t  
tan  
t t
t
t
t
3
lim
3
lim
3 lim
= t  
= t  
=
t  
1
1
1
Dt  
Dt  
t
t
t
5
5

15 lim sec 2   = 15 sec 2  lim
 = 15 sec 2 0  15
t  
t



t
t

Thus,
Thus,
t
t
lim ln y =
 
lim y =
 
t
5
lim 3 t tan    15
 
t
3t
lim ( e )
t  
5
tan  
 t 
= e 15
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Answer: e 15
7.
x6
lim ( e )
 a 
sin  6 
 x 
x  
, where a is a constant
 a 
sin  6 
6
 a 
e   and lim sin  6  = sin 0  0 , then lim ( e x )  x 
Since xlim
x  
x  
 
x 
0
=   Indeterminate Form.
x6
Let y  ( e )
x6
 a 
sin  6 
 x 
 a 
 a 
x6
6
. Then ln y  sin  6  ln e = x sin  6  .
x 
x 
 a 
x 6 sin  6  , which has an indeterminate form of   0 .
Now, find xlim
 
x 
 a 
sin  6 
x 
 a 
x 6 sin  6  = xlim
Since xlim
, which has an indeterminate form
 
1
 
x 
x6
0
of , then we can applying L’Hopital’s Rule to this last limit. Thus,
0
 a   a 
 a 
 1 
 a 
cos  6  D x  6 
cos  6  a D x  6 
sin  6 
x  x 
x 
x 
x 
lim
lim
lim
= x
= x
=
x  
1
 1 
 1 
D
D




x
x
6
6
x6
x 
x 
a 
 a 

a lim cos  6  = a cos  lim 6  = a cos 0  a ( 1)  a
x  
x 
x x 
 a 
6
ln y = lim x sin  6   a
Thus, xlim
x  
 
x 
x6
y = lim ( e )
Thus, xlim
 
x  
 a 
sin  6 
 x 
= ea
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Answer: e a
g ( x )  L and lim h( x )  L and if
Theorem (The Sandwich Theorem) If xlim
 a
x  a
g ( x )  f ( x )  h( x ) for all x in a deleted neighborhood of x  a , then
lim f ( x )  L .
x  a
f ( x )  L , we need to show that
Proof Let   0 . In order to show that xlim
 a
there exists a   0 such that f ( x)  L   whenever 0  x  a   .
Let { x : 0  x  a   a } = ( a   a , a )  ( a , a   a ) be a deleted neighborhood
of x  a for which g ( x )  f ( x )  h( x ) . Thus, g ( x )  f ( x )  h( x ) whenever
0  x  a  a
g ( x )  L , then there exists a  g  0 such that g ( x)  L  
Since xlim
 a
whenever 0  x  a   g . Since g ( x)  L       g ( x )  L   
L    g ( x )  L   , then L    g ( x )  L   whenever 0  x  a   g .
Thus, L    g (x ) whenever 0  x  a   g .
h( x )  L , then there exists a  h  0 such that h( x)  L  
Since xlim
 a
whenever 0  x  a   h . Since h( x)  L       h( x )  L   
L    h( x )  L   , then L    h( x )  L   whenever 0  x  a   h .
Thus, h( x )  L   whenever 0  x  a   h .
Let  = min { a ,  g ,  h } . Then    a ,    g , and    h .
Since    a , then { x : 0  x  a   }  { x : 0  x  a   a } . Since
g ( x )  f ( x )  h( x ) whenever 0  x  a   a , then g ( x )  f ( x )  h( x )
whenever 0  x  a   .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Since    g , then { x : 0  x  a   }  { x : 0  x  a   g } . Since
L    g (x ) whenever 0  x  a   g , then L    g (x ) whenever
0 x  a .
Since    h , then { x : 0  x  a   }  { x : 0  x  a   h } . Since
h( x )  L   whenever 0  x  a   h , then h( x )  L   whenever
0 x  a .
Thus, L    g ( x )  f ( x )  h( x )  L   whenever 0  x  a   . Thus,
L    f ( x )  L   whenever 0  x  a   . Since L    f ( x )  L  
f ( x )  L   . Thus,
0  x  a   . Thus, lim f ( x )  L .
x  a
    f ( x)  L   
f ( x )  L   whenever
COMMENT: Some people use the phrase Squeeze Theorem instead of Sandwich
Theorem.
Examples Find the following limits, if they exist.
1.
lim
x  
sin x
x
Since  1  sin x  1 for all x and x  0 (since x is approaching positive
1 sin x 1
 for all x  0 .
infinity), then  
x
x
x
1
sin x
 1
   0 and lim
 0 , then lim
 0 by the

Since xlim
 
x   x
x  
x
 x
Sandwich Theorem.
Answer: 0
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
2.
lim
cos 2 t
3t2
lim
cos 2 t
1
cos 2 t
lim
=
2
3t
3 t   t2
t  
t  
2
Since  1  cos 2 t  1 for all t and t  0 for t approaching negative
1
cos 2 t
1

infinity, then  2 
for all t  0 .
t
t2
t2
 1 
Since t lim    2   0 and
 t 
Sandwich Theorem.
Thus,
lim
t  
lim
t  
1
 0 , then
t2
lim
t  
cos 2 t
 0 by the
t2
cos 2 t
1
cos 2 t
1
lim
(0)  0
=
=
2
3t
3 t   t2
3
Answer: 0
3.
lim
x  
lim
x  
 3
 5 sin 
 x

7x3




 3
 5 sin 
 x

7x3






sin
NOTE:


3
x
 3
sin 
 x
5


lim
=
7 x  
x3





 is defined for all x  0 and is undefined if x  0 .


Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

Since  1  sin 


  1 for all x  0 and x 3  0


 3
sin 
 x
1

approaching negative infinity), then  3 
x
x3
 3 

sin 

 x 
1
1




for all x  0 .
x3
x3
x3
3
x
1
 1 
 0 and lim   3   0 , then
3
x  
x
 x 
the Sandwich Theorem.
Since xlim
 
Thus,
lim
x  
 3
 5 sin 
 x

7x3




(since x is



  1 
x3

sin 


lim
x  
 3
sin 
 x
5

=  x lim



7
x3




3
x
x3
= 



 0
by
5
(0)  0
7
Answer: 0
4.
4
cos 2  
 
lim
  
5
4
2 4 
NOTE: cos   and hence cos   is defined for all   0 .
 
 
4
2 4 
Since  1  cos    1 for all   0 , then 0  cos    1 . Since
 
 
4
cos 2  
   1
 5  0 for  approaching positive infinity, then 0 
for all
5
5
  0.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
4
cos 2  
1
   0
0  0 and lim 5  0 , then lim
Since  lim
by the
   
  
 
5
Sandwich Theorem.
Answer: 0
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860