Download - Limits (L'Hospital's Rule)
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Limits Indeterminate Forms and LHospitals Rule
I. Indeterminate Form of the Type
0
0
We have previously studied limits with the indeterminate form0
0as shown in the
following examples:
Example 1: ( ) 42222
)2)(2(
2
4limlimlim
22
2
2
=+=+=
+=
x
x
xx
x
x
xxx
Example 2: === xx
x
x
xx
x
x
xxx 2sin
1
3cos
1
1
3sin
2sin
3cos3sin
2sin
3tanlimlimlim
000
2
3)1)(1)(1(
2
3
2sin
2
3cos
1
3
3sin
2
3limlimlim
02003
==
x
x
xx
x
xxx
[Note: We use the given limit 1sin
lim0
=
.]
Example 3:12
1
83
1)8(
28
3 2
3
0lim ===
+
fh
h
h. [Note: We use the definition
of the derivativeh
afhafaf
h
)()()( lim
0
+=
where 3)( xxf =
and a = 8.]
Example 4: ( ) ( ) 23
3sin
3
3
21cos
lim3
===
fx
x
x
. [Note: We use the
definition of the derivativeax
afxfaf
ax
=
)()()( lim where
xxf cos)( = and 3=a .]
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However, there is a general, systematic method for determining limits with the
indeterminate form0
0. Suppose thatfandgare differentiable functions atx = a
and that)(
)(lim
xg
xf
axis an indeterminate form of the type
0
0; that is, 0)(lim =
xf
ax
and 0)(lim =
xgax
. Sincefandgare differentiable functions atx = a, thenfandg
are continuous atx = a; that is, )()( lim xfafax
= = 0 and )()( lim xgagax
= = 0.
Furthermore, sincefandgare differentiable functions atx = a, then = )(af
ax
afxf
ax
)()(lim and
ax
agxgag
ax
=
)()()( lim . Thus, if 0)( ag , then
)(
)(
)(
)(
)()(
)()(
)()(
)()(
)(
)(limlimlimlim
xg
xf
ag
af
ax
agxgax
afxf
agxg
afxf
xg
xf
axaxaxax
=
=
=
=
iff and
g are continuous atx = a. This illustrates a special case of the technique known as LHospitals Rule.
LHospitals Rule for Form0
0
Suppose thatfandgare differentiable functions on an open interval
containingx = a, except possibly atx = a, and that 0)(lim =
xfax
and
0)(lim = xgax . If )()(
lim xg
xf
ax
has a finite limit, or if this limit is + or
, then)(
)(
)(
)(limlim
xg
xf
xg
xf
axax
=
. Moreover, this statement is also true
in the case of a limit as ,,, + xaxax or as .+x
In the following examples, we will use the following three-step process:
Step 1. Check that the limit of)(
)(
xg
xfis an indeterminate form of type
00 . If it
is not, then LHospitals Rule cannot be used.
Step 2. Differentiatefandgseparately. [Note: Do not differentiate)(
)(
xg
xf
using the quotient rule!]
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Step 3. Find the limit of)(
)(
xg
xf
. If this limit is finite, + , or , then it is
equal to the limit of)(
)(
xg
xf. If the limit is an indeterminate form of type
00 , then simplify )(
)(
xg
xf
algebraically and apply LHospitals Rule again.
Example 1: 4)2(21
2
2
4limlim
2
2
2
===
x
x
x
xx
Example 2:2
3
)1(2
)1(3
2cos2
3sec3
2sin
3tan 2
00limlim === x
x
x
x
xx
Example 3:121
)8(3
1
)8(3
11
)1()8(
3
1
283
23
20
32
0
3
0limlimlim ==
+=
+=+
h
h
hh
hhh
Example 4: ( )2
3
3sin
1
sin
3
21cos
limlim33
==
=
x
x
x
xx
Example 5:2
1
22
11limlimlim
002
0
==
=
x
x
x
x
x
x
e
x
e
x
xe[Use LHospitals Rule
twice.]
Example 6: ( ) ( )( ) ( ) 010
1cos
2
11cos
2
1sin
1
limlimlim2
32
===
=
+++x
x
xx
x
x
x
xxx
, or
( )0
1
)0(2
cos
2
sin1sin
1
limlimlim0
2
0
2
====++ + y
y
y
y
x
x
yyx
where xy 1= .
Example 7: ( ) 0)0(0ln1
ln limlim 00 === xxxx
xx [This limit is not an indeterminate
form of the type0
0, so LHospitals Rule cannot be used.]
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II. Indeterminate Form of the Type
We have previously studied limits with the indeterminate form
as shown in the
following examples:
Example 1: =
+
+=
+
+
++222
2
222
2
2
2
132
753
132
753limlim
xx
x
x
x
xx
x
x
x
xx
xx
xx
2
3
002
003
132
753
limlim
2
2=
+
+=
+
+
++ xxxx
xx
Example 2: =+
1
13
2lim x
x
x =
+
22
2
22
1
13
lim
xx
x
xx
x
x 0
1
0
01
00
11
13
2
2
lim ==+
=+
x
xx
x
Example 3: =+
12
43
2
3
limx
x
x
=
+
33
2
33
3
12
43
lim
xx
x
xxx
x =
+
=+
0
3
00
03
12
43
3
3
lim
xx
x
x
limit does not exist.
Example 4: =++
1
14 2
limx
x
x
=+
+
x
xx
x
x 1
14 2
lim
x
x
x
x
x 1
14
2
2
lim +
+
xx =2(
becausex < 0 and thus 2xx = ) = =+
+
x
x
x
x
x 1
14
2
2
lim
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21
4
11
14
2
2
lim =
=+
+
x
x
x
.
However, we could use another version ofLHospitals Rule.
LHospitals Rule for Form
Suppose thatfandgare differentiable functions on an open interval
containingx = a, except possibly atx = a, and that =
)(lim xfax
and
= )(lim xgax . If )( )(lim xg xfax has a finite limit, or if this limit is + or
, then)(
)(
)(
)(limlim
xg
xf
xg
xf
axax
=
. Moreover, this statement is also true
in the case of a limit as ,,, + xaxax or as .+x
Example 1:2
3
4
6
34
56
132
753limlimlim 2
2
==+
=++
+++ xxx x
x
xx
xx
Example 2: 0)0(2
31
2
3
2
3
1
13limlimlim 2 ====+
xxx
x
xxx
Example 3: ===+
4
18
4
9
12
43limlimlim
2
2
3 x
x
x
x
x
xxx
Example 4: +
=+=++
144
1142
8
114
2
22
limlimlimxxx
x
xx
xxx
LHospitals
Rule does not help in this situation. We would find the limit as we
did previously.
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Example 5:( )( ) 24
4
22
3
3
2
2
3
2
33
22
13
12
1
3
1
2
)1ln(
)1ln(limlimlimlim
xx
xx
xx
xx
x
x
x
x
x
x
xxxx ++
=++
=
+
+=++
++++=
3
2
72
48
72
48
636
24
612
28
limlimlim 2
2
3
3
===+=++
+++ x
x
x
x
xx
x
xxx
Example 6: 02
0
222
1
1
ln 22
0
3
03
02
0limlimlimlim =
=
=
=
=++++
x
x
x
x
x
x
x
xxxx
Example 7: 02
)0(arctan1arctan
limlimlim =
=
=
+++
x
xx
x
xxx
[This limit is
not an indeterminate form of the type
, so LHospitals Rule
cannot be used.]
III. Indeterminate Form of the Type 0
Indeterminate forms of the type 0 can sometimes be evaluated by rewriting theproduct as a quotient, and then applying LHospitals Rule for the indeterminate
forms of type
0
0or
.
Example 1: 0)(1
1
1
lnln limlimlimlimlim
0
2
02
000
==
=
==+++++
xx
x
x
x
x
xxx
xxxxx
Example 2: =
=
==++++ x
xx
xx
x
x
xxx
xxxx
tansin
cotcsc
1
csc
lnln)(sin limlimlimlim
0000
0)0)(1(tan
sin
limlim 00 ==
++ xx
x
xx
Example 3: ( )(
1sin
1
1sin1sin limlimlim
0
===+++ y
y
x
xx
xyxx
[Let xy 1= .]
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IV. Indeterminate Form of the Type
A limit problem that leads to one of the expressions
)()( ++ , )()( , )()( ++ , )()( ++
is called an indeterminate form of type . Such limits are indeterminatebecause the two terms exert conflicting influences on the expression; one pushes
it in the positive direction and the other pushes it in the negative direction. However,
limits problems that lead to one the expressions
)()( +++ , )()( + , )()( + , )()( +
are not indeterminate, since the two terms work together (the first two produce alimit of + and the last two produce a limit of ). Indeterminate forms of thetype can sometimes be evaluated by combining the terms and manipulating
the result to produce an indeterminate form of type00 or
.
Example 1: =+
=
=
+++ xxx
x
xx
xx
xx xxx sincos
1cos
sin
sin
sin
11limlimlim
000
02
0
coscossin
sinlim
0
==++
+ xxxx
x
x
Example 2: ( )[ ] =
=
2
0
2
0
cos1lnln)cos1ln( limlim
x
xxx
xx
=
=
2
1ln
2
sinln
cos1ln limlim
02
0 x
x
x
x
xx
V. Indeterminate Forms of the Types 1,,0 00
Limits of the form [ ])(
)(limxg
ax
xf
[ ]
)()(lim
xg
x
xfor frequently give rise to
indeterminate forms of the types 1,,0 00 . These indeterminate forms can
sometimes be evaluated as follows:
(1) [ ] )()( xgxfy =
(2) [ ] [ ])(ln)()(lnln )( xfxgxfy xg ==
(3) [ ] [ ]{ })(ln)(ln limlim xfxgyaxax
=
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The limit on the righthand side of the equation will usually be an
indeterminate limit of the type 0 . Evaluate this limit using the
technique previously described. Assume that [ ]{ })(ln)(lim xfxgax
=L.
(4) Finally, [ ] L
axaxax
eyLyLy ==
=
limlimlim lnln .
Example 1: Findx
x
xlim0+
.
This is an indeterminate form of the type 00 . Let === xx xyxy lnln
xx ln . ( ) =====
+++++ x
x
x
x
xxxy
xxxxxlimlimlimlimlim
02
0000 1
1
1
lnlnln 0.
Thus, 10
0lim ==
+exx
x.
Example 2: Findxx
x
e2
)1(lim
++ .
This is an indeterminate form of the type 0 . Let +=
xxey2
)1(
x
eey
x
xx)1ln(2
)1(lnln2 +
=
+=
.
x
ey
x
xx
)1ln(2ln limlim
+=
++=
22
1
2
1
12
limlimlim =
=+
=
+
+++x
x
xx
x
x
x
x
x e
e
e
ee
e
. Thus,xx
x
e2
)1(lim
++ =
2e .
Example 3: Find ( ) xx
x1
0
coslim+
.
This is an indeterminate form of the type 1 . Let = xxy1
)(cos
x
xxy x
)ln(cos)(coslnln
1
=
= . ==
++ x
xy
xx
)ln(cosln limlim
00
( ) 0tanlim0
=+x
x. Thus, ( ) x
x
x 1
0
coslim+
= 10 =e .
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Practice Sheet for LHospitals Rule
(1) =
)2cos(1
3
0lim
x
xxex
x
(2) =+ 3)(ln
limx
x
x
(3) =
)]ln()cos1[ln( 2
0lim xxx
(4) =
+
x
x x
32
1lim
(5) =
+ 12
cos
11
cos
lim
x
x
x
(6) =
x
xx
x
2
0
11lim
(7) =
2
1
0
)(coslim xx
x
(8) =+
++
1683
275
234
234
1lim
xxx
xxxx
x
(9) =
x
x
x
5819lim
0
(10)( ) =++ 15ln
2ln3
3
limxx
x
(11) ( ) =+
+
xx
x
e2
1lim
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(12) =
30
)2sin(2)4sin(lim
x
xx
x
(13) =
xx
x
1
sin
1lim
0
(14) =
+1
1
lim xx
ex
(15) =+
xxx
ln3
0lim
(16) =
++
x
x
x
x
15
12ln
lim0
(17) =
+
+
21lim
x
x x
e
(18) = x
x
x
)3arctan(sinlim
0
(19)( )
=+
3
0
2
0
sin
limx
dttx
x
(20) ( ) =+
xx
x
xe1
2
0lim
(21) =+ x
x
x
arctanlim
(22) = )tan2arcsin())3(arctan(sin
lim0 x
x
x
(23) = 20
)ln(coslim
x
x
x
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(24) =
+
+
x
x x2
11lim
(25) =+
x
x
x1
)(lnlim
(26) ( =++
xex
x
)1ln(lim
Solution Key for LHospitals Rule
(1) 2
3
4
6
2cos4
339
2sin2
13
)2cos(1
333
0
33
0
3
0 limlimlim ==++
=+
=
x
eexe
x
exe
x
xxexxx
x
xx
x
x
x
(2) ( ) ( )=====
+++++ x
x
xxx
x
xxx
x
xxxxx ln61ln6
1
)(ln31)(ln3
1
)(lnlimlimlimlimlim 223
( )+==
++ 616
1limlim
x
xxx
(3) =
=
=
2
0
2
0
2
0
cos1ln
cos1ln)]ln()cos1[ln( limlimlim
x
x
x
xxx
xxx
2ln2
1ln
2
sinln lim
0
=
=
x
x
x
(4) Let =x
y1
( ) yy
x
x
yx
3
0
3
212
1 limlim =
++. Now, let ( ) == zyz y ln21
3
( )( )
61
21
6
)21ln(3ln
21ln321ln limlimlim
000
3
=
=
=
=+++
y
y
yz
y
yy
yyy
y. Thus,
==
=
+++
6
000limlimlim 6ln6ln ezzzyyy
( ) yy
x
x
yx
3
0
3
2121 limlim =
++=
6
0lim
=
+ez
y.
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(5) Let =x
y1 =
=
=
++ + )2sin(2
)sin(
1)2cos(
1)cos(
12
cos
11
cos
limlimlim00 y
y
y
y
x
x
yyx
41
cos41
cossin4sin limlim
00== ++ yyy
yyy
.
(6) =+
+
=
2
22
0
2
0 11
111111limlim
xx
xx
x
xx
x
xx
xx
( ) ( )( ) ( ) 2
1
11
1
1111
11
202
2
02
2
0limlimlim =
+
=
+
=
+
xx
x
xxx
xx
xxx
xx
xxx.
(7) Let ( ) ===== 200211
)ln(cosln)ln(cos)ln(cosln)(cos limlim22
xxy
xxxyxy
xx
xx
2
1
cos2
1sin
cos2
sin
2
cos
sin
limlimlim000
=
=
=
xx
x
xx
x
x
x
x
xxx
. Thus, ( ) = 2
1lnlim
0
yx
==
21
00limlim
2
1ln eyy
xx
21
0
1
0limlim
2
)(cos
== eyx
x
x
x
.
(8) =+
+=
+
+=
+
++
124836
24260
122412
122120
1683
2752
2
123
23
1234
234
1
limlimlimxx
xx
xxx
xxx
xxx
xxxx
xxx
0
20limit does not exist.
(9) ( )=
+
=++
=
xx
x
x
x
x
x
x
x
xxx 5819
)581(81
5819
581958195819limlimlim
000
( ) 185
5819
5
5819
5limlim
00
=+
=+ xxx
x
xx.
(10)( )( )
13015
315
)2(15
)15(3
15
15
2
3
15ln
2ln3
3
3
3
3
2
3
2
3
3
limlimlimlim =+
=+
=
+=+
++++ x
x
x
x
x
x
x
x
x
x
xxxx
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(11) Let ( ) ( ) =+=+=+=+
y
x
eeyey
x
x
xxxx ln)1ln(2
1lnln1 lim22
2
2
1
2
1
1
2
)1ln(2limlimlimlim =
=
+
=+
=+
++++x
x
xx
x
x
x
x
x
x
x e
e
e
ee
e
x
e . Thus, =+
yx
lnlim
==
++
2limlim 2ln2 eyyxx
( ) 22
limlim 1
+
+==+ eye
x
xx
x.
(12) =+
=
=
x
xx
x
xx
x
xx
xxx 6
)2sin(8)4sin(16
3
)2cos(4)4cos(4)2sin(2)4sin(limlimlim
02
03
0
86
48
6
)2cos(16)4cos(64lim
0
=
=+
xx
x
.
(13) =+
=
=
xxx
x
xx
xx
xx xxx sincos
cos1
sin
sin1
sin
1limlimlim
000
02
0
coscossin
sinlim
0
==++ xxxx
x
x
.
(14)(
.11
1
1
11 0
1
2
2
11
1
limlimlimlim ===
=
=
++++ee
x
xe
x
eex x
x
x
x
x
x
x
x
(15) .0)3(3
31
1ln
ln 3
0
34
034
031
0
3
0limlimlimlimlim ==
=
==
+++++ x
x
x
x
x
x
xxx
xxxxx
(16)3
)15)(12(
3
1
)15(
)12(5)15(2
12
15
15
12ln
limlimlim0
2
00
=++
=
+
++
++
=
++
xx
x
xx
x
x
x
x
x
xxx
.
(17) Let =x
y 1 ( ) yy
x
x
eyx
e2
1
0
2
11 limlim +=
+
++. Next, let ( ) =+= zeyz y ln1 2
1
( )
22
1
2
)1ln(ln
2
)1ln(1ln limlimlim
000
21 eey
e
y
eyz
y
eyey
yyy
y =+
=+
=+
=++++
. Thus,
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==
=
+++
2
000limlimlim
2ln
2ln
e
yyy
eze
ze
z ( ) yy
x
x
eyx
e2
1
0
2
11 limlim +=
+
++=
2
0
lime
y
ez=+
. ( ) ( ) ( ) =+=+=+=
y
x
xexeyxey
x
x
xxxx lnln
lnln lim0
21
21
2
(18)3
1
3sin1
3cos3
)3arctan(sin 2
00limlim =
+=
x
x
x
x
xx
.
(19)( )
3
1
3
)cos(
6
)cos(2
3
)sin(sin 2
0
2
02
2
03
0
2
0limlimlimlim ====
++++
x
x
xx
x
x
x
dtt
xxx
x
x
.
(20) Let ( ) ( ) ( ) =+=+=+=
yx
xexeyxeyx
x
xxxx lnlnlnln lim0
2
1212
( )3
1
12
ln 2
2
0
2
0limlim =+
+
=+
xe
e
x
xe x
x
x
x
x
. Thus =
=
3ln3ln limlim
00
yyxx
=
3
0lim eyx
( ) 30
12
0limlim eyxex
xx
x
==+
.
(21)0
2arctan
lim =+=+
x
x
x
.
(22)2
3
121
3
tan41
sec2
3sin1
3cos3
)tan2arcsin(
))3(arctan(sin
2
2
2
00limlim ==
+=
x
x
x
x
x
x
xx.
(23)
2
1
cos2
1sin
2
cos
sin
)ln(coslimlimlim
002
0
=
=
= xx
x
x
x
x
x
x
xxx
.
(24) Let =x
y1
.2
11
2
11
1
0limlim
y
y
x
x
yx
+=
+
++Let =
+= zyz
y
ln2
11
1
14
-
7/30/2019 Limits (L'Hospital's Rule)
15/15
.
2
1
1
211
21
2
11ln
ln2
11ln
2
11ln limlimlim
000
1
=+
=
+
=
+
=
+
+++
y
y
y
zy
y
yyyy
y
Thus, ==
=
+++
21
000limlimlim
2
1ln
2
1ln ezzz
yyy
=
+
+
x
x x2
11lim
.2
11 2
1
0
1
0limlim ezyy
y
y
==
+
++
(25) Let =====++ x
xy
x
xxyxy
xx
xx)ln(ln
ln)ln(ln
)ln(lnln)(ln limlim11
0
ln
1
1
ln1
limlim ==++ xx
xx
xx
. Thus, ==
=
+++yyy
xxxlimlimlim 0ln0ln
=10e 1)(ln limlim1
==++
yxx
x
x.
(26) ( ) ( )( ) =
+=+=+
+++x
x
x
xx
x
x
x e
eeexe
1lnln1ln)1ln( limlimlim
01lnln1
ln limlim ==
=
+
++x
x
xx
x
x e
e
e
e.
15