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INDETERMINATE FORMS
OBJECTIVES:• define, determine, enumerate the
different indeterminate forms of functions;
• apply the theorems on differentiation in evaluating limits of indeterminate forms of functions using L’Hopital’s Rule.
.
2313-x lim1x
3x1-x lim1x
3x4x lim
:follows as numerator the factor we exist, to limit the for and form, ateindetermin an is limit the
00
113)1(4)1(
1x3x4x lim
1x
3x4x lim of itlim the Evaluate :callRe
1x1x
2
1x
22
1x
2
1x
21x
3x4x lim ,thus2
1x
used. be will Rule sHopital'L' on Theorems limit said the evaluate To example. second the to applied be longerno can problems previous the in applied principle the Obviously,
00
0)0sin(
)0(2)0(2sin
2x2x sin lim
2x
2x sin lim the evaluating consider us Let
0x
0x
.
1 , ,0 5. and - 4. 0 3.
:Forms Secondary B.
2.
and 00 1.
:Forms Primary A.:forms ateindeterminof Kinds
00
.
Theorem 3.6.1 (p. 220) L'Hôpital's Rule for Form 0/0
.
Applying L'Hôpital's Rule (p. 220)
Theorem 3.6.2 (p. 222) L'Hopital's Rule for Form /
.
2x2x sin lim .1
0x
EXAMPLE:Evaluate the following limits.
00
00sin
0202sin
2x2x sin lim
0x
10 cos2
02cos2
122x2coslim
2xdxd
2x sindxd
lim2x
2x sin lim
:Rule s'Hopital'L gsinu By
0x0x0x
12x
2x sin lim 0x
.3y sin-y3y-y tan lim .2
0y
0
0 0000
0sin3-003-0tan
3y sin-y3y-y tan lim
0y
122
3131
0 3cos-130 sec
33y cos-113ysec lim
3y sin-ydxd
3y-y tandxd
lim3y sin-y3y-y tan lim
:LHR By
2
2
0y0y0y
13y sin-y3y-y tan lim
0y
.
2
4x x4
2x sin ln lim .3
4x42
2x2cos2x sin
1
limx4
dxd
2x sin lndxd
limx42x sin ln lim
:LHR By
4x2
4x
2
4x
.ateminerdetin still is This
00
082
2cot
448
42 cot 2
x482x 2cot lim
4x
0
00
2sinln
44
42 sinln
x42x sin ln lim 22
4x
01 ln:Note
ln 0 ln
.
32x2csc4 lim
)4(82x2csc2 lim
x48dxd
2x cot 2dxd
lim
:LHR peatRe
2
4x
2
4x
4x
811
81
42 csc
81x2csc
81 lim 2
2
2
4x
8
1x4
sin2x ln lim 2
4x
.
x
2
x ex lim .4
ee
x lim2
x
2
x
e
21e
2x lime
dxd
xdxd
limex lim
:LHR By
xxx
2
xx
2
x
02e2
1e12lim
edxd
2xdxd
lim
:LHR peatRe
xxx
x
0ex lim x
2
x
.3x tan ln3x cos ln lim .5
6x
- ln
0 ln
2 tan ln
2 cos ln
63 tan ln
63 cos ln
3x tan ln3x cos ln lim
6x
01 ln:Note
ln 0 ln
3x3secx3tan
1
3x3sincos3x
1
lim3x tan ln
dxd
3x cos lndxd
lim3x tan ln3x cos ln lim
:LHR Apply
26x
6x
6x
x3cos1
3xcosx3sin
limx3secx3tan lim
3x3secx3tan
13x tan3 lim
2
2
2
6x
2
2
6x26
x
16
3sinx3sin lim2
2
6x
13x tan ln3x cos ln lim
6x
.
applies. Rule sHopital'L' case theof
either In . or 00 to result may which evaluated is limit the then
xg1xf lim xgxf lim
, Hence one. equivalent an to dtransforme is products their limit, such evaluate To limit. its approaches
x as 0 or 0 form the having undefined is xg and xf
of product the unsigned), or signed be could (which 0xg lim and
0xf lim that such functions abledifferenti two are xg and xfIf .A:onDefininiti
axax
ax
ax
- and 0 FORMS ATEINDETERMIN The
.
Rule. sHopital'L' apply Then . or 00 to result may
evaluated when limit whose quotient equivalent an into difference the ngtransformi by evaluated be could limit The
. xg lim xf lim xgxf lim is That .-
form theof ateindetermin be to said is xgxf lim the
then , positive both are which xg lim and , xf limIf .B
axaxax
ax
axax
- and 0 FORMS ATEINDETERMIN The
.
LHR. then and logarithmof properties the apply then function, the for y variable
a letting by evaluated be may forms ateindetermin Thesely.respective ,1 , ,0 forms ateminerdetin the assumed
xf lim expression the then or approaches x as or
xg lim and ,1xf lim
or ,0xg lim and ,xf lim
or ,0xg lim and ,0xf lim :if and ,xg and xf functions two Given
:Definition
00
xg
ax
axax
axax
axax
1 and , ,0 FORMS ATEINDETERMIN The 00
.
EXAMPLE:Evaluate the following limits:
2x csc x lim .10x
00csc02x csc x lim0x
00
0sin0
sin2xx lim2x csc x lim
:function rational equivalent an to function the gminTransfor
0x0x
212x csc xlim
21
)0(cos21
2cos2x1 lim
2cos2x1 lim
sin2xdxd
xdxd
limsin2x
x lim
:LHR Apply
0x
0x0x0x0x
.
x ln x lim .20x
00 ln 0x ln x lim0x
010 ln
x1x ln limx ln x lim
:function equivalent an to function given the gminTransfor
0x0x
0x ln x lim 0x
0xlim
x1
1x1
lim
x1
dxd
x lndxd
lim
x1x ln lim
:LHR Apply
0x
2
0x0x0x
.
1x1
x ln1 lim .3
1x
01
01
111
1 ln1
1x1
x ln1 lim
1x
0
01 ln 1-11 ln11
x ln 1-xx ln1xlim
1x1
x ln1 lim
:fraction simple a ot gminTransfor
1x1x
1x ln1x11x
1x11
limx ln 1-x
dxd
x ln1xdxd
lim1x
1x ln
1 lim
:LHR Apply
1x1x1x
00
1 ln 11111
x ln x1-x1x lim
xx ln x1-x
x1-x
lim1x1x
.
1xln1x1x1
1 limx ln x1-x
dxd
1-xdxd
limx ln x1-x
1-x lim
:LHR again Apply
1x1x1x
21
1 ln21
x ln21 lim
1x
21
1x1
x ln1 lim
1x
.
x2secx1
x 1 lim .4
220x
-
01
01
0sec01
01
x2secx1
x1 lim
220x
00
002cos1
xcos2x-1 lim
xx2cos
x1 lim
x2secx1
x1 lim
: function equivalent the to gminTransfor
20x220x220x
00
002sin
x22x2sinlim
xdxd
cos2x-1dxd
limx
cos2x-1 lim
:LHR Apply
0x2
0x20x
20cos2x2cos2lim
112x2coslim
xdxd
x2sindxd
limx
x2sinlim
:again LHR Apply
0x0x0x0x
2x2secx
1x1 lim 220x
.
x
0xx2 lim .5
00x
0x00 2x2 lim
xx2y Let
x1
x2 ln2x ln xy ln
x2 lny ln x
0ln
01
02ln
x1
x2ln limyln lim
:sides both on itlim the Apply
0x0x
0xlim
x1
2x2
1
x1
dxd
x2lndxd
lim
x1
x2lnlim
:LHR Apply
0x
2
0x0x
12x lim therefore then
2xy since
1y limey lim:sides bothof function inverse the Take
x
0x
x
0x
0
0x
0yln lim
0
x1
x2ln limyln lim
0x
0x0x
.
1x1
1xxlim .6
111xlim 0
111
11x
1
1x
1x1
xy Let
1x
xlnxln1x
1x lny ln 1x1
00
111ln
1xxlnlimyln lim
:1x as sides both on itlim the Applying
1x1x
11
x1lim
1
1x1
lim1x
dxd
xlndxd
lim1x
xlnlim
:member right the on LHR Apply
1x1x1x1x
.
72.2exlim
xy but eylim
:sides bothof function inverse the take ,1ylnlim1x
xlnlim
,Thus
1x1
1x
1x1
1
1x
1x1x
x
0xxcotlim .7
00x
0x0cotxcotlim
x1
xcot lnxcot ln xxcot lny ln
xcoty Let
x
x
ln
01
0 cot lnlim
x1
xcotlnlimylnlim
:sides both on limit the pplyA
0x
0x0x
2
2
0x0x0x
x1
1xcscxcot
1
lim
x1
dxd
xcotlndxd
lim
x1
xcotlnlim
:member right on LHR Apply
xcosxsin2x2lim
x1
xcosxsin1
lim
x1
xsin1
xcosxsin
lim2
0x2
0x2
2
0x
0
00sin
02x2sin
x2lim22
0x
x2cosx2lim
2x2cosx4lim
x2sindxd
x2dxd
limx2sin
x2lim
:again LHR Apply
0x0x
2
0x
2
0x
010
0cos02
1x cotlim then x coty Since
1eylim
sides bothof function inverse the take ,0yln lim
x1
xcotlnlim
,Hence
x
0x
x
0
0x
0x0x
x4sinxtanx2lim .1
0x
220y y1
ysin1lim.2
xsinx2lim.3 10x
ycosln
ylim.42
0y
x3x2lnlim.5
3
x
x2tan1
x1ln1lim.8 10x
x4
2
0xx1lim.9
x2
2
x ex3lim.10
2x2
2
0xxsin1lim.11
x2
x
0xx3elim.12
x2tanlnx2coslnlim.13
4x x
1
0xx2sinx2coslim.6
xcscxsinlim.15 1
0x
x
x2
0xe1lim.7
xlnxcoslim.14 1
0x
EXERCISES: Evaluate the following limits.