More Related Content Similar to L16 indeterminate forms (l'hopital's rule) Similar to L16 indeterminate forms (l'hopital's rule) (20) More from James Tagara (6) 2. 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.
3. .
( )( )
( ) 2313-xlim
1x
3x1-x
lim
1x
3x4x
lim
:followsasnumeratorthefactorweexist,tolimit
theforandform,ateindeterminanislimitthe
0
0
11
3)1(4)1(
1x
3x4x
lim
1x
3x4x
limofitlimtheEvaluate:callRe
1x1x
2
1x
22
1x
2
1x
−=−==
−
−
=
−
+−
=
−
+−
=
−
+−
−
+−
→→→
→
→
2
1x
3x4x
lim,thus
2
1x
−=
−
+−
→
used.bewillRulesHopital'L'onTheoremslimit
saidtheevaluateToexample.secondthetoappliedbelonger
nocanproblemsprevioustheinappliedprincipletheObviously,
0
0
0
)0sin(
)0(2
)0(2sin
2x
2xsin
lim
2x
2xsin
limtheevaluatingconsiderusLet
0x
0x
===→
→
8. .
2x
2xsin
lim.1 0x→
EXAMPLE:
Evaluate the following limits.
( )
( )
( )
0
0
0
0sin
02
02sin
2x
2xsin
lim0x
===→
( )
( )
( )
( )
( )
10cos
2
02cos2
12
2x2cos
lim
2x
dx
d
2xsin
dx
d
lim
2x
2xsin
lim
:Rules'Hopital'LgsinuBy
0x0x0x
===
== →→→
1
2x
2xsin
lim
0x
=∴
→
9. .3ysin-y
3y-ytan
lim.2 0y→
( ) ( )
( ) ( ) 0
0
00
00
0sin3-0
03-0tan
3ysin-y
3y-ytan
lim0y
=
−
−
==→
( )
( )
( )
( )
( )
1
2
2
31
31
03cos-1
30sec
33ycos-1
13ysec
lim
3ysin-y
dx
d
3y-ytan
dx
d
lim
3ysin-y
3y-ytan
lim
:LHRBy
2
2
0y0y0y
=
−
−
=
−
−
=
−
=
−
== →→→
1
3ysin-y
3y-ytan
lim
0y
=∴
→
10. .
( )
( )2
4
x x4
2xsinln
lim.3
−ππ
→
( )
( )
( )
( )
( )
( )( )4x42
2x2cos
2xsin
1
lim
x4
dx
d
2xsinln
dx
d
lim
x4
2xsinln
lim
:LHRBy
4
x2
4
x
2
4
x −−π
=
−π
=
−π π
→
π
→
π
→
( ) ( )
.ateminerdetinstillisThis
0
0
08
2
2cot
4
48
4
2cot2
x48
2x2cot
lim
4
x
−
π
=
π
−π−
π
=
−π−π
→
( )
( ) 0
0
0
2
sinln
4
4
4
2sinln
x4
2xsinln
lim 22
4
x
=
π
=
π
−π
π
=
−ππ
→
( ) 01ln
:Note
=
( ) ∞=∞ln
( ) −∞=0ln
11. .
[ ]
( )[ ]
( )
32
x2csc4
lim
)4(8
2x2csc2
lim
x48
dx
d
2xcot2
dx
d
lim
:LHRpeatRe
2
4
x
2
4
x
4
x
−
=
−−
−
=
−π−
π
→
π
→
π
→
( )
8
1
1
8
1
4
2csc
8
1
x2csc
8
1
lim
2
2
2
4
x
−=−=
π
−=−⇒ π
→
( )
( ) 8
1
x4
sin2xln
lim 2
4
x
−=
−
∴
→ ππ
12. .
x
2
x
e
x
lim.4 +∞→
( )
∞+
∞
=
∞+
=⇒ ∞++∞→
ee
x
lim
2
x
2
x
[ ]
[ ] ( )
( )
∞+
∞+
=
∞+
=== ∞++∞→+∞→+∞→
e
2
1e
2x
lim
e
dx
d
x
dx
d
lim
e
x
lim
:LHRBy
xx
x
2
xx
2
x
[ ]
[ ]
( )
( )
0
2
e
2
1e
12
lim
e
dx
d
2x
dx
d
lim
:LHRpeatRe
xx
x
x
=
∞+
====⇒ ∞++∞→+∞→
0
e
x
lim x
2
x
=∴
+∞→
13. .
3xtanln
3xcosln
lim.5
6
x
π
→
( )
( ) ∞
∞
=
∞
=
π
π
=
π
π
=⇒ π
→
-
ln
0ln
2
tanln
2
cosln
6
3tanln
6
3cosln
3xtanln
3xcosln
lim
6
x
( ) 01ln
:Note
=
( ) ∞=∞ln
( ) −∞=0ln
[ ]
[ ]
( )
( )3x3sec
x3tan
1
3x3sin
cos3x
1
lim
3xtanln
dx
d
3xcosln
dx
d
lim
3xtanln
3xcosln
lim
:LHRApply
2
6
x
6
x
6
x
−
== π
→
π
→
π
→
x3cos
1
3xcos
x3sin
lim
x3sec
x3tan
lim
3x3sec
x3tan
1
3xtan3
lim
2
2
2
6
x
2
2
6
x26
x
π
→
π
→
π
→
−=
−=
•
−
( ) 1
6
3sinx3sinlim
2
2
6
x
−=
π
−=⇒ π
→
1
3xtanln
3xcosln
lim
6
x
−=∴
→
π
14. .
( ) ( ) ( )
( )
( ) ( )
( ) ( )
( )
( )
applies.RulesHopital'L'casetheof
eitherIn.or
0
0
toresultmaywhichevaluatedislimitthethen
xg
1
xf
limxgxflim
,Henceone.equivalentantodtransforme
isproductstheirlimit,suchevaluateTolimit.itsapproaches
xas0or0formthehavingundefinedisxgandxf
ofproducttheunsigned),orsignedbecould(which0xglimand
0xflimthatsuchfunctionsabledifferentitwoarexgandxfIf.A
:onDefininiti
axax
ax
ax
∞
∞
=•
•∞∞•
=
=
→→
→
→
∞∞∞• -and0FORMSATEINDETERMINThe
15. .
( ) ( ) ( )
( ) ( )[ ]
( ) ( )[ ] ( ) ( )
Rule.sHopital'L'applyThen.or
0
0
toresultmay
evaluatedwhenlimitwhosequotientequivalentaninto
differencethengtransformibyevaluatedbecouldlimitThe
.xglimxflimxgxflimisThat.-
formtheofateindeterminbetosaidisxgxflimthe
then,positivebotharewhichxglimand,xflimIf.B
axaxax
ax
axax
∞
∞
∞−∞=−=−⇒∞∞
−
∞=∞=
→→→
→
→→
∞∞∞• -and0FORMSATEINDETERMINThe
16. .
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )[ ] ( )
LHR.thenandlogarithm
ofpropertiestheapplythenfunction,theforyvariable
alettingbyevaluatedbemayformsateindeterminThese
ly.respective,1,,0formsateminerdetintheassumed
xflimexpressionthethenorapproachesxasor
xglimand,1xflim
or,0xglimand,xflim
or,0xglimand,0xflim
:ifand,xgandxffunctionstwoGiven
:Definition
00
xg
ax
axax
axax
axax
∞
→
→→
→→
→→
∞
∞−∞+
∞==•
=∞=•
==•
∞
∞ 1and,,0FORMSATEINDETERMINThe 00
17. .
EXAMPLE:
Evaluate the following limits:
[ ]2xcscxlim.1 0x→
[ ] [ ] ∞•==→
00csc02xcscxlim0x
[ ]
0
0
0sin
0
sin2x
x
lim2xcscxlim
:functionrationalequivalentantofunctionthegminTransfor
0x0x
=== →→
[ ]
[ ] ( )( )
[ ]
2
1
2xcscxlim
2
1
)0(cos2
1
2cos2x
1
lim
2cos2x
1
lim
sin2x
dx
d
x
dx
d
lim
sin2x
x
lim
:LHRApply
0x
0x0x0x0x
=∴⇒==
===
→
→→→→
18. .
[ ]xlnxlim.2 0x→
[ ] ( ) ( )∞−==→
00ln0xlnxlim0x
[ ]
∞
∞−
=== →→
0
1
0ln
x
1
xln
limxlnxlim
:functionequivalentantofunctiongiventhegminTransfor
0x0x
[ ] 0xlnxlim
0x
=∴
→
[ ] ( )
( ) 0xlim
x
1
1
x
1
lim
x
1
dx
d
xln
dx
d
lim
x
1
xln
lim
:LHRApply
0x
2
0x0x0x
=−=
−
=
= →→→→
20. .
( ) [ ]
[ ] ( ) ( )( )1xln1
x
1
x1
1
lim
xlnx1-x
dx
d
1-x
dx
d
lim
xlnx1-x
1-x
lim
:LHRagainApply
1x1x1x
++
=
+
=
+
⇒ →→→
( ) 2
1
1ln2
1
xln2
1
lim1x
=
+
=
+
⇒ →
2
1
1x
1
xln
1
lim
1x
=
−
−∴
→
21. .
−→
x2secx
1
x
1
lim.4 220x
( )
∞∞=−=−=
−→
-
0
1
0
1
0sec0
1
0
1
x2secx
1
x
1
lim 220x
( )
0
0
0
02cos1
x
cos2x-1
lim
x
x2cos
x
1
lim
x2secx
1
x
1
lim
:functionequivalentthetogminTransfor
20x220x220x
=
−
=
=
−=
− →→→
[ ]
[ ]
( )( ) ( )
0
0
0
02sin
x2
2x2sin
lim
x
dx
d
cos2x-1
dx
d
lim
x
cos2x-1
lim
:LHRApply
0x
2
0x20x
==
−−
==
→→→
( ) ( )
( )
( )( )( )
20cos2x2cos2lim
1
12x2cos
lim
x
dx
d
x2sin
dx
d
lim
x
x2sin
lim
:againLHRApply
0x0x0x0x
===== →→→→
2
x2secx
1
x
1
lim 220x
=
−∴
→
22. .
[ ]x
0x
x2lim.5 →
[ ] ( )[ ] 00x
0x
002x2lim ==→
[ ]x
x2yLet =
[ ]
[ ]
x
1
x2ln
2xlnxyln
x2lnyln
x
==
=
( )
∞
∞−
=
∞
=== →→
0ln
0
1
02ln
x
1
x2ln
limylnlim
:sidesbothonitlimtheApply
0x0x
[ ] ( )
( ) 0xlim
x
1
2
x2
1
x
1
dx
d
x2ln
dx
d
lim
x
1
x2ln
lim
:LHRApply
0x
2
0x0x
=−=
−
=
= →→→
( )
( ) 12xlimthereforethen
2xysince
1ylimeylim
:sidesbothoffunctioninversetheTake
x
0x
x
0x
0
0x
=
=
=→=
→
→→
0ylnlim
0
x
1
x2ln
limylnlim
0x
0x0x
=
==
→
→→
23. .
( ) 1x
1
1x
xlim.6 −
→ +
( ) ( ) ( ) ( )∞
−−
→
===+
111xlim 0
1
11
1
1x
1
1x
( ) 1x
1
xyLet −=
( ) ( )
1x
xln
xln
1x
1
xlnyln 1x
1
−
=
−
== −
0
0
11
1ln
1x
xln
limylnlim
:1xassidesbothonitlimtheApplying
1x1x
=
−
=
−
=
→
++
→→
+
( )
( )
( )
1
1
x
1
lim
1
1
x
1
lim
1x
dx
d
xln
dx
d
lim
1x
xln
lim
:memberrighttheonLHRApply
1x1x1x1x
===
−
=
− ++++
→→→→
24. .
( )
( ) 72.2exlim
xybuteylim
:sidesbothoffunctioninversethetake,1ylnlim
1x
xln
lim
,Thus
1x
1
1x
1x
1
1
1x
1x1x
==∴
==
==
−
−
→
−
→
→→
+
+
++
( )x
0x
xcotlim.7 +
→
( ) ( ) 00x
0x
0cotxcotlim ∞==+
→
( )
( )
x
1
xcotln
xcotlnxxcotlnyln
xcotyLet
x
x
===
=
25. ( ) ( )
∞
∞
=
∞
∞
=
=
+
++
→
→→
ln
0
1
0cotln
lim
x
1
xcotln
limylnlim
:sidesbothonlimitthepplyA
0x
0x0x
( ) ( )( )
2
2
0x0x0x
x
1
1xcsc
xcot
1
lim
x
1
dx
d
xcotln
dx
d
lim
x
1
xcotln
lim
:memberrightonLHRApply
−
−
=
= +++
→→→
xcosxsin2
x2
lim
x
1
xcosxsin
1
lim
x
1
xsin
1
xcos
xsin
lim
2
0x
2
0x
2
2
0x ⋅
⋅
==
= +++
→→→
( )
( ) 0
0
0sin
02
x2sin
x2
lim
22
0x
=== +
→
26. ( )
( ) ( )( ) x2cos
x2
lim
2x2cos
x4
lim
x2sin
dx
d
x2
dx
d
lim
x2sin
x2
lim
:againLHRApply
0x0x
2
0x
2
0x ++++
→→→→
===
( )
( )
0
1
0
0cos
02
===
( ) ( ) 1xcotlimthenxcotySince
1eylim
sidesbothoffunctioninversethetake,0ylnlim
x
1
xcotln
lim
,Hence
x
0x
x
0
0x
0x0x
=∴=
==
==
+
+
++
→
→
→→
27. x4sin
xtanx2
lim.1
0x
+
→
−
→ 220y y
1
ysin
1
lim.2
xsin
x2
lim.3 10x −→
→ ycosln
y
lim.4
2
0y
( )
x3
x2ln
lim.5
3
x +∞→
( )
−
+ −→ x2tan
1
x1ln
1
lim.8 10x
( )x
4
2
0x
x1lim.9 +
→
+∞→ x2
2
x e
x3
lim.10
( ) 2
x
2
2
0x
xsin1lim.11 +
→
( )x
2
x
0x
x3elim.12 +
→
x2tanln
x2cosln
lim.13
4
x
π
→( )x
1
0x
x2sinx2coslim.6 −
→
( )( )xcscxsinlim.15 1
0x
−
→
x
x
2
0x
e1lim.7
++
→
( ) xlnxcoslim.14 1
0x
−
→ +
EXERCISES: Evaluate the following limits.
