Mat h 12 06 Ca lc ul us – Sec . 4. 6: Indet er mi nate F or m s a nd L’H o pita l’s R ul e
I.
Int r od ucti o n
A. Review
1. Evaluate the following.
a.
# x " 2&
lim %
(
x !2 $ x " 3'
b.
" ex %
d. lim+ $$ ''
x !0 # x &
" sin x $
c. lim
x! 0 # x %
e.
# x"3&
lim % 2
(
x ! 3$ x " 9 '
" ln x %
lim+ $
'
x !0 # cot x &
2. Note: b, c and e are limits of an indeterminate form. In Chapter 2, we
encountered such limits and used algebra manipulation, geometry or tables in order
to find the limit. In this section we will learn to apply L’Hopital’s rule to evaluate
limits of indeterminate forms.
B. Indeterminate and Determinate Forms
1. Indeterminate Forms
When we have equations whose limits involve a form that looks like:
0 !
,
, 0 " ! , ! # ! , 00 , !0 , 1! , we have a form that is indeterminate. i.e. we
0 !
are not sure if there is a limit or not.
2. Determinate Forms:
The following are determinate forms:
II.
! + ! " ! , - ! # ! " #! ,
0 ! " 0 , 0 -! " !
L’H op ital’ s R ul e
A. Background
John Bernoulli discovered a rule for calculating limits of fractions whose numerators
and denominators both approach zero. This rule is today known as L'Hopital's rule after
Guillaume Francois Antoine de l'Hopital (1661-1704) who published the first calculus
text in 1696.
B. Theorems:
1. L’Hopital’s Rule (First Form): Suppose that f(a)=g(a)=0, that f '(a) and g '(a) exist,
and that
g !(a ) " 0 . Then lim
x! a
f ( x)
g ( x)
=
f "( a )
g " (a )
.
2. L'Hopital's Rule (Strong Form): Suppose that f and g are differentiable on an
open interval I containing a. Suppose also that
g !( x ) " 0 on I if x ! a . Suppose
that
lim f ( x ) = 0 and lim g( x ) = 0
x! a
x! a
or
that
lim f ( x ) ! ±" and lim g( x ) ! ±" .
x! a
(In other words, we have an indeterminate form of type
lim
x! a
f ( x)
g ( x)
= lim
x !a
f "( x )
g "( x )
x! a
0
!
.) Then
or
0
!
, i f the limit on the right exists (or is
! or " ! ).
NOTE: In the strong form it is important that you realize something. We do not
know that the first limit is actually equal to the second limit, UNTIL we have shown
that the second limit exists or is infinite. In other words if the second limit can not
be established then we can not equate the two limits.
III.
Ex am pl es
A. Indeterminate Quotients:
0
!
, ±
0
!
Evaluate the following limits.
1.
# x"3&
lim % 2
(
x ! 3$ x " 9 '
2.
" sin x $
lim #
x! 0
x %
3.
" ln x %
lim+ $
'
x !0 # cot x &
4.
# y2 &
((
lim %%
y !" $ ln y '
5.
$ e# x '
)
lim &&
#x )
x !" % 1 + e
(
6.
# 1 " e 3t + 3t &
((
lim %%
2
t !0 $
t
'
7.
# y"4 &
((
lim %% 3
y !4 $ y + 4 " 2 '
B. Other Indeterminate Forms
1.
Indeterminate Products and Differences:
0! " , " #"
0 ! " , " # " then you will have to use
0
!
algebra to manipulate the equation so that you get either
.
or
0
!
a. If you have an indeterminate form
b. Examples: Evaluate the following limits.
1.)
lim x (e x # 1)
1
x !"
2.)
3.)
lim (! cot ! )
! "0+
# 1
1&
lim+ %
" 2(
t !0 $ sin t
t '
2. Indeterminate Powers:
0
0
0
a. Limits that lead to 0 ,
function in terms of e.
lim
x !a
[ f (x )] ( ) =
g x
!
0 , ! , 1
0
!
! , 1 can sometimes be handled by rewriting the
( ) ln f ( x )
lim e g x
x! a
lim g ( x ) ln f ( x )
= e x! a
b. Examples: Evaluate the following limits.
1.)
lim + x 3x
x!0
2.)
3.)
lim (1+ x ) x
1
x !0+
( x# )
lim# (tan x )
x! "2
"
2