L’Hospital’s Rule

Lesson 4.5

Problem

• There are times when we need to evaluate functions which are rational

• At a specific point it may evaluate to an indeterminate form

3

2

27( )

9

xf x

x

001 0

0

Example of the Problem

• Consider the following limit:

• We end up with the indeterminate form

• Note why this is indeterminate

3

23

27lim

9x

x

x

0

0

00 0 ?

0n n n

L’Hospital’s Rule

• When gives an indeterminate

form (and the limit exists)– It is possible to find a limit by

• Note: this only works when the original limit gives an indeterminate form

( )lim

( )x c

f x

g x

'( )lim

'( )x c

f x

g x

001 0

0

Example

• Consider

As it stands this could be

• So we claim

2

2

1 2 1lim lim 1

2 1x x

xx x x

x

2limx

x x x

Example

• Consider

• Why is this not a candidate for l’Hospital’s rule?

• Note also example 7, pg 232 .. the limit must exist

0

1 coslim

secx

x

x

0

1 cos 0lim

sec 1x

x

x

This is not an

indeterminate result

This is not an indeterminate result

Example

• Try

• When we apply l’Hospital’s rule we get

• We must apply the rule a second time

20

1 coslimx

x

x

0

sinlim

2x

x

x

Hints

• Manipulate the expression until you get one of the forms

• Express the function as a fraction to get

0 001 0 0

0

( )

( )

f x

g x

Assignment

• Lesson 4.5

• Page 236

• Exercises 1 – 55 EOO