Calculus and Analytical Geometry

Lecture # 11

MTH 104

L’HÔPITAL’S RULE: INDETERMINATE FORMS

INDETERMINATE FORMS OF TYPE

0 0

( )lim

( )x a

f x

g x

( ) 0f x x a

0 0

Limit of the form

in which and

is called an indeterminate form of type

.

( ) 0 as g x

L’Hopital Rule for form 0/0

f g

x a x a

lim ( ) 0 lim ( ) 0x a x a

f x and g x

Suppose that and are differentiable functions on an open

, except possible at , and that

interval containing

lim ( ) / ( )x a

f x g x

If exists, or if this limit is or , then

( )lim

( )x a

f x

g x

( )lim

( )x a

f x

g x

Moreover this statement is also true in the case of limits as

, , ,x a x a x ,x or as

EXAMPLE Find the limit 2

2

4lim

2x

x

x

Using L’HÔPITAL’S rule, and check the result by factoring.

solution2 2

2

4 2 4lim

2 2 2x

x

x

0

0

form

Using L’HÔPITAL’S rule

2

2

4lim

2x

dx

dxdx

dx

2

2

4lim

2x

x

x

2

2lim

1x

x

4

By computation2

2

4lim

2x

x

x

2

( 2)( 2)lim

( 2)x

x x

x

4 2

lim 2x

x

ExampleIn each part confirm that the limit is an indeterminate form of type 0/0 and evaluate it using L’HOPITAL’s rule

0

sin 2limx

x

x / 2

1 sinlim

cosx

x

x

(a)

(b)

30

1lim

x

x

e

x

2

0

tanlimx

x

x

(c)

(d)

20

1 coslimx

x

x

4 / 3lim

sin(1/ )x

x

x

(e)

(f)

Solution

(a)0

sin 2limx

x

x

sin 0

0%

Applying L’HÔPITAL’S rule

0

sin 2limx

x

x

0

(sin 2 )lim

( )x

dx

dxdx

dx

0

2cos 2lim 2cos(0)

1x

x

2

(b)

/ 2

1 sinlim

cosx

x

x

1 sin 2cos 2

1 1 000

Applying L’HÔPITAL’S rule

/ 2

1 sinlim

cosx

x

x

/ 2

1 sinlim

cosx

dx

dxd

xdx

/ 2

coslim

sinx

x

x

cos 2sin 2

00

1

(c)30

1lim

x

x

e

x

0 1

0

e 1 1

0

0

0

Applying L’HÔPITAL’S rule

30

1lim

x

x

e

x

20lim

3

x

x

e

x

0

0

e 1

0

20

tanlimx

x

x

tan 0

0 0

0

(d)

Applying L’HÔPITAL’S rule

20

tanlimx

x

x

20

tanlimx

dx

dxdx

dx

2

0

seclim

2x

x

x

INDETERMINATE FORMS OF TYPE

( ) / ( )f x g x

The Limit of a ratio, in which the numerator has limit and the denominator has the limit is called an indetreminate form of type L’Hopital Rule for form Suppose f and g are differentiable functions on an open interval conatining x=a, except possibly at, x=a and that

lim ( )x a

f x

and lim ( )x a

g x

lim ( ) ( )x a

f x g x

If exists, or if this limit is or , then

( ) ( )lim lim

( ) ( )x a x a

f x f x

g x g x

Moreover this statement is also true in the case of limits as

, , ,x a x a x ,x or as

Example

In each part confirm that the limit is indeterminate form of type

and apply L’HÔPITAL’S

limxx

x

e 0

lnlim

cscx

x

x(a) (b)

solution

limxx

x

e e

(a)

Applying L’HÔPITAL’S rule

limxx

x

e

( )lim

( )x x

dx

dxde

dx

1lim

xx e 1 0

0

lnlim

cscx

x

x

ln 0

csc(0)

(b)

Applying L’HÔPITAL’S rule

0

lnlim

cscx

x

x

0

(ln )lim

(csc )x

dx

dxd

xdx

0

1lim

csc cotx

x

x x

1 x csc x cot xAny additional application of L’HÔPITAL’S rule will yield powers of

in the numerator and expressions involving and

in the denominator.

Rewriting last expression

0

sinlim tanx

xx

x

0

sinlimx

x

x

0lim tanx

x

(1)(0) 0

Thus, 0

lnlim 0

cscx

x

x

INDETERMINATE FORMS OF TYPE0.

The limit of an expression that has one of the forms

( )( ), ( ). ( ), ( ) , ( ) ( ), ( ) ( )

( )g xf x

f x g x f x f x g x f x g xg x

( )f x ( )g xis called and indeterminate form if the limits andindividually exert conflicting influences on the limit of the entire expression.

For example

0lim ln 0.x

x x

On the other hand

2lim 1x

x x

( )

Indeterminate form

Not an indeterminate form

0.

0 / 0

Indeterminate form of type

can sometimes be evaluated by rewriting the product as a ratio, and then applying L’HÔPITAL’S rule for indeterminate form of type or .

Example Evaluate

0lim lnx

x x

(a) (b)/ 4

lim (1 tan )sec2x

x x

0lim lnx

x x

0.( )

Solution

(a)

0lim lnx

x x

0

lnlim

1x

x

x

Rewriting form

Applying L’HÔPITAL’S rule

0lim lnx

x x

0

lnlim

1x

x

x

0

(ln )lim

1( )x

dx

dxd

xdx

0 2

1lim

1x

x

x

2

0limx

x

x

0lim ( )x

x

0

/ 4lim (1 tan )sec2x

x x

(b)

0

Rewriting as

4

lim 1 tan sec2x

x x

4

1 tanlim

1 sec2x

x

x

4

1 tanlim

cos 2x

x

x

0

0

Applying L’HÔPITAL’S rule

4

lim 1 tan sec2x

x x

4

1 tanlim

cos 2x

x

x

2

4

seclim

2sin 2x

x

x

2(sec )42

2sin4

2

12

4

1 tanlim

cos2x

dx

dxd

xdx

Indeterminate forms of type

A limit problem that leads to one of the expressions

,

,

is called an indeterminate form type

.

The limit problems that lead to one of the expressions

,

,

are not indeterminate, since two terms work together.

Example Evaluate

0

1 1lim

sinx x x

Solution

0

1 1lim

sinx x x

1 10 sin 0

0

1 1lim

sinx x x

0

sin

sinlimx

x x

x x

sin 0 00sin 0

Rewriting

00

Applying L’HÔPITAL’S rule

0

1 1

sinlimx x x

0

sin

sinlimx

x x

x x

0

sin

sinlimx

dx x

dxdx x

dx

0

cos 1

sin coslimx

x

x x x

0

0

Again Applying L’HÔPITAL’S rule

0

cos 1lim

sin cosx

dx

dxd

x x xdx

0

1 1

sinlimx x x

0

sinlim

cos sin cosx

x

x x x x

sin 0

cos0 0sin 0 cos0

0

01 0 1

INDETERMINATE FORM OF TYPE 0 00 , ,1

( )lim ( ) xf x g

can give rise to indeterminate forms of the types and 0 00 , 1

1

0lim (1 ) x

xx

It is indeterminate because the expressions and gives --Two conflicting influences. Such inderminate form can be evaluated by first introducing a dependent variable

( )( )g xy f x

1 and respectively.

( )1n 1n ( )

g xy f x

( ) 1n ( )g x f x

The limit of lny will be an indeterminate form of type 0

Limits of the form

For example

(1 form)

1 x1x

Example1

0lim (1 ) xx

x e

Solution Let 1(1 ) xy x

1

ln ln(1 ) xy x 1

ln ln(1 )y xx

0 0

ln(1 )lim ln limx x

xy

x

0( form)0ln(1 0)

0

Applying L’HÔPITAL’S rule

0 0

1limln lim 1

1x xy

x

0 0

11lim ln lim1x x

xy

ln 1 as 0y x

ln 1 as 0ye e x as 0.y e x

1

0Thus, lim 1 .x

xx e