Exponential Indeterminate Forms: The Logarithm Method
(For 11, 1 0, and 00 Types)
Dave L. Renfro
The Logarithm Method: Justi…cation
Recall the (sequential) de…nition of continuity: If () is continuous at = 
, then
lim () = (
). Since = lim (), we can recast this into the following form:
!
!
³
´
lim [ (
) ] =  lim () 
!
!
More generally, suppose that is a function of and is a function of . Furthermore, assume
that lim+ () exists (we’re not assuming that () is continuous on the right at = 
, or that
!
(
) even exists) and that () is continuous at = lim+ (). Then
!
µ
¶
1
lim+ (()) =  lim+ () 
!
!
. Since the logarithm of 
 is easier to handle, we
Let’s make use of this to evaluate lim+ 
!0
introduce a logarithm:
lim+ [ ln () ] =
!0
lim+ [ ¢ ln () ] = lim+
!0
!0
·
¸
ln (
)
=
¡1
l’Hôpital
lim+
!0
·
¡1
¡¡2
¸
= 0
0
Therefore (exponentiate both sides), lim+  = 
= 1. To see exactly how this is justi…ed, let
!0

() = 
, () = ¢ ln (
), and = 0. Then lim+ [¢ ln (
)] = 0 (by l’Hôpital; see above),
µ
¶
µ
¶!0
so  lim+ () becomes exp lim+ [¢ ln ()] = exp (0) = 1. This is supposed to equal
!
!0
lim+ (()) = lim+ [exp (¢ ln ())] = lim+ [exp (ln ())] = lim+ [],2 which happens to be
!
!0
!0
the limit being sought.
!0
The Logarithm Method: Worked Examples
1. lim [1 + 2sin ()]cot() .

!0
2 sin()]
Let = [1 + 2sin ()]cot(). Then we have ln () = cot () ¢ ln [1 + 2sin ()] = ln[ 1 +
.
tan()
Hence,
2
3
·
¸
2 cos()
·
¸
ln [1 + 2 sin ()]
2 cos3 ()
1+2 sin()
=
5 = lim
lim ln () = lim
lim 4
= 2
!0
!0
l’Hôpital !0
!0
tan ()
sec2 ()
1 + 2sin ()
Exponentiating both sides gives
h
i
exp lim ln (
)
!0
1
2
=
c o nt i n ui t y of
 at  = 2
lim [exp (ln )] = lim =
!0
!0
2


Of co ur se, the ana log ous version for ! ¡ als o ho lds.
The …rst equa lity co mes fro m plugg ing in fo r , , and . The la st two equalities a re a lgebraic ma nipula tions.
¡
¢

  (
2. lim + 
, 
, and are …xed constants)

!0
¡
¢
 
Let = + 
. Then ln (
)=
lim ln () = lim
!0
!0
"

¢
¡
¢


ln 
+ 
=
¡
¢#


¢ ln + 
=
l’Hôpital4

¡
¢

 
Exponentiating both sides gives lim 
+ 

!0
³
sin()

Let =
³
´ 12
si n()

´
1
2
2
lim ln () = lim 4
!0
!0
. Then ln (
)=
ln
³
s in()

2

´3
5
·
1
2
¢ ln
³
sin( )

=
l’Hôpital
,3 which gives
2 ¢ + 3
(
)
+
4
5 = (+ 
lim
)
!0
1
(+ 
)

.
=
!0
3. lim

¢ln(+)

´
, which gives
lim
!0
"
¢cos() ¡s in()
2
¥
2
sin()

#
= lim
!0
·
¢ cos () ¡ sin (
)
22 ¢ sin ()
¸
¸
·
¸
·
¸
2 ¢ cos (
¡¢ sin ()
4¢ sin () + 2
) ¡1
4¢ sin () + 22 ¢ cos () ¡1
5
lim
= lim
=
lim
! 0
!0
l’Hôpital !0 4¢ sin () + 22 ¢ cos ()
¡¢ sin ()
¡¢ sin ()
·
µ
¶¸¡1

1
= lim ¡4 ¡ 2 ¢ cos ¢
= [¡4 ¡ 2 ¢ 1 ¢ 1]¡1 = ¡ 
!0
sin (
)
6
³
´ 12
¡
¢

)
Exponentiating both sides gives lim sin(
=
exp ¡ 16 .

=
!0
Some Additional Limits (With Answers!) to Practice With6
¡
¢1
lim 1 + 2  = 1

!0
lim
!1+
lim
·
lim
h
!0
!0
 ¡
1¡+ln()
³
lim+ 
!0
i
= ¡2
si n(2 ) ¡si nh(2 )
6
¸
¢sin(s in ) ¡si n2 ()
6
lim
!0
= ¡ 13
i
=
1
18
lim
!0
h
h
´

( )
1 ¡ cos()
3¡2
=1
i
sin(¢cos )
¢sin()
1
lim [ln (1 + )]= 1
¡2
lim [cos (2)] 2 = 
!0
!0+
lim+ ( ) = 0
h
1
1
lim ln() = 
!1
lim+
!0
=0
i
=
lim
!0

2
lim [1 + 
]  = 1 ( 0)
!1
lim
!0
h
h
£1
+
!0+
¤
ln () = 1
tan()

i
1
¢s in()
1
2
¡
1
2
lim [1 + sin (2
)]  = 
!0
1
3
=
1
2
1
i

lim [1 + 
] = 
!0
=
lim
·
cos( 
))
2 ¢cos(
sin 2( )
lim
h
3 ¢sin()
(1 ¡ cos )2
!0
1
6
!0
lim
!0
h
sin()

i
i
¸
! 0
!0
a ct ua lly co rrect fo r all choices of 
,
, and 
.
5
6
¡1 at = ¡6 jus ti…es go ing fr om lim [stu¤ ] ¡1 to
Cont inuity of () = 
So me of these involve
0, 1,
0 1
!0
0 ¢ 1, or 1 ¡ 1 indeterminate for ms.
·
¸ ¡1
lim (stu¤ )
.
!0

4
=4

¡ sin()
¡
¢
 is de…ned. In other wor ds , yo u want t o
Check t o see that
for the
+ 
¡
¢ purpos es of the pr es ent limit , ln 


m ake s ur e that ln +  is de…ned for all values o f s u¢ciently clo se t o 0. [Consider  0, consider the e¤ect
t ha t one or more of t he …xed co ns tants being neg ative (o r zer o) has, etc.]
4 If = 0, or if = = 0, then this isn’t an indeterminat e lim it. H owever, in either of these cases it is immediate
(+). T herefore, this answer is
t ha t lim ln () = 0, result ing in lim = 1. N ote this agr ees wit h our answer 
3
=
¡1
=