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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 )
ba
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
= x0
= 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 t0
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
sin 2 x
lim
=
x 0
5x
1
5
lim 4
x 0
1
5
sin 2 x
lim
=
x 0
x
x cos 2 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
11
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
11
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
0
x 0
x 0
8x
4x
8x
4 x0 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 x0 x
4 x0
4
2
Answer:
6.
9
2
lim
1 cos t
t3
lim
1 cos t
1 cos 0
11
0
Indeterminate Form
=
=
=
t3
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 t0 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 t0 t
6 t0
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 t0 t
6 t0
t
1
sin t
lim 2 .
L’Hopital’s Rule, then
3 t0 t
Thus,
lim
Thus,
lim
t 0
1 cos t
1
sin t
lim 2 .
=
3
t
3 t0 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
sec 2 6 x
and
Thus, x lim
0
x 2/3
+
+
0
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
11
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
2 ( 4 w 5) 2 4
w
w
w
w
w
1
12 8 w 3 w
w 10
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 2 4
w
w
w
w
003
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
x 4
lim
x 4
5 x 11 3
x 2
9 3
5 x 11 3
x 2
=
4 2
=
0
Indeterminate Form
0
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 ) 2 / 3
3 ( 24 )
3 (6)
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
21
5 x 2 19 4
=
3
21
64 4
=
0
Indeterminate Form
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
011
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
303
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 x0
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
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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 t0
2t
3 t0
t
3 t 0 t2
2
sin 8 t
lim
3 t0 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 t0 t
3 t0
3 t0
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 t0
2t
3 t0 t
16
16
lim cos 8 t =
3 t0
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
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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
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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
=
x2
lim
x
lim
x
=
lim
x
lim
x
e 5x
. The
x2
x2
e 5x
. The
e 5x
, we have that
x2
5e 5 x
5
5e 5 x
= x lim
2x 3
2 x3
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
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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
x
x
x
x
5
5e
5
e
25
e
25
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
=
5x
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
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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
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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
=
3x0
x x 0
3 x0
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
3x0
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
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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
lim
x
sin
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
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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
x5
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
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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
=
70
600
60
=
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
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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
lim
y
(0) 0
=
=
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
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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
lim
9
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
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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
Dx
1
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
Dt
1
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
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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
= e8
1
e8
lim x x
x 0
lim x x = 0 0 Indeterminate Form.
x 0
ln x
Let y x . Then ln y x ln x = 1 .
x
x
Now, find
lim
x 0
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
= x lim
7
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
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