indeterminate forms and l’hopital’s rule lesson 8.7
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Indeterminate Forms and L’Hopital’s Rule
Lesson 8.7
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’Hopital’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
• Must change to format
• So we manipulate algebraically and proceed
2limx
x x x
'( )lim
'( )x c
f x
g x
2 2
2
2lim limx x
x x x x x xx x x
x x x
Example
• Consider
• Why is this not a candidate for l’Hospital’s rule?
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 8.7
• Page 574
• Exercises 1 – 33 EOO
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