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Transcript of Limit Calculus)
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Limits
Limits are used to describe how a function behave as the independent variable move toward
a certain value. Consider a funtion 1,111)(
2{!
! xx
xxxf and the tables below .
Table Table
l)(l
!
pp x
xxf
xx
!
pp xxxf
xx
2
1lim
1
11lim
1
1
!
!
!
p
p
x
x
xx
x
x
2
1lim
1
11lim
1
1
!
!
!
p
p
x
x
xx
x
x
Since lili !!
pp
xfxfxx
, therefore !p
xx
.
The function is undefined at !x but the limit does exist.
x xf
. .
.0 .0
,00 .00
.000 .000
.0000 .0000
x xf
0.9 .9
0.99 .99
0.999 .999
0.9999 .9999
0.99999 .99999
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Example
!
"
!
3if7
3if4
3if4
xx
x
xx
xf
27lim)(lim 233
!! ppxxf
xx
242lim)(lim33
!!
pp
xxfxx
!
p
xfx
Example
lim)(lim00
!!
pp
xxfxx
g!!
pp xxf
xx
2li)(li
00
istnotoes)(li0
xfxp
y
x
y
x
e
"!
0if
0if)(
xx
xxxf
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Example
!x
x
g!
!
p
p
p 3li
li
li
3
3
3 x
xf
x
x
x
!
!
p
p
p 3lim
lim
lim
3
3
3 xxf
x
x
x
istnotoes)(li xfx p
Example
!
xxf
g!
!
p
p
p 2
2
2
2
2l
1l
l
x
xf
x
x
x
g!
!
p
p
p2
2
2
2 2lim
1lim
)(limx
xf
x
x
x
g!
)(li2
xfx
y
x-3
y
x2
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Exa pl { l xfx g
and l xfx g
}
1l m !g
xfx
4)(lim
!gpx
fx
Exa pl { l xfx g and l xfx g }
6)(lim !gp
xfx
g!gp
)(lim xfx
-1
y
x
4
-3
y
x
6
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Exa pl { )(li xfx gp
and )(li xfx gp
}
g
gp
)(li xfx
g!gp
)(lim xx
L t Co putat onal t chn qu s
kkcx!
p
lim , wh r k sa constant.
kkkxx
!!gpgp
limlim
cxcx!
lim
-3
y
x2 6
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Example 3lim3
!
xx
0li0
!
xx
2li2
p
xx
g!gxxli g!
gx
xlim
g!p xx
1l g!p xx
1l xistnotdoes1lim0 xxp
i3
!
xx
2
11li2
! xx
0lim !p xx
0lim !!p xx
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Theorem
Let m stand for oneof the m ts
c"
p
li ,
pcx
lim ,
pcx
lim ,
gpx
lim or
gpx
lim . If )(lim Lxf ! and
2li Lxf ! bothexist, then
a limlimlim LLxgxfxgxf !! .
b (i(i((i LLxgxfxgxf !! .
c 21)(i)(i)()(i LLxgxfxgxf !! .
d if,(li
(li
(
(li 2
2
1 {!!
L
L
L
xg
xf
xg
xf.
e eveinifovi e(li(li 11 "!! LLxfxfnnn .
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Example
a)
03)3(33limlim4lim34lim
2
33
2
3
2
3
!!! pppp xxxx
xxxx
b)
8
24
4)4(5
2li
45li
245
li
4
4
4!
!
!
p
p
p x
x
xx
x
x
x
c) 42li
2
22li
24li
22
2
2!!
!
ppp
xx
xx
xx
xxx
d)
exist.otdoes12
1li
2
p xxx
x
g!
# 12
1li
2 xx
x
x
g!
$ 12
1li
2 xx
x
x
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E mp Find
a)28
li3
2
p xx
x b)
11li
p x
x
x
c) )(li xfxp
if
u!
3if2)1(3
3if13)(
2xx
xxxf
Solution
a) 1222lim
2
222lim
2
8lim 22
2
22
2
%
2
!!
!
&
&
&
xxx
xxx
x
x
xxx
b) 211
li11
11
11li
11li
000!!
v
! ppp x
xx
x
x
x
x
x
x
xxx
c) 11)()1(li)(li33
!!!
''
xxfxx
102)2(32)1(3lim)(lim22
33 !!! (( xxf xx
10li3
!@)
xfx
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Example Find
a) )(li2
xfx p
if
"
e!
2if134
2if32)(
2 xxx
xxxf
b) a if
"
e!
2if2
2if1-)(
2
3
xax
xxxf and )(lim
2xf
xpexists.
Solution a) 91)2(3)2(4)134(lim)(lim22
22!!!
ppxxxf
xx
228)2(3232lim)(lim 22 !!!! pp xxf xx )(lim
2xf
x p@ does not exist.
b) )(lim2
xfxp
exists )(lim)(lim22
xfxfxx
00
!
4
5
54
1)2(2)2(
)1(lim)2(lim
32
3
2
2
2
!
!
!
! 11
a
a
a
xax xx
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Limitsfortrigonometricfunctions
Letcis real number
a) cxcx
sinsinlim !p
b) cxcx
coscosli !p
c) cxcx
tantanli !p
d) cxcx
csccsclim !p
e) cxcx
secseclim !p
f) cxcx
cotcotli !p
ample
a) 12
iili
2
!T!T
2
x
x
b)4
4
14
1
14
4tan
1tanli
4
T!
T
!
T
T
!T
px
x
x
c) 2
4cos
1cos
1limcossin
coscossin
limcossin
1tanlim
444
!T
!!
!
T3
T3
T3
xxxx
xx
xxx
xxx
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Speci l Li its
I. 1silim !p x
x
x
II. 0cos1lim0
!p x
x
x
Example
a
)1
cos
1
li
sin
li
cos
sin
li
tan
li !!!
pppp xx
x
xx
x
x
x
xxxx
b) 44
4sinlim44
44sinlim4sinlim !!v!444
xx
xx
xx
xxx
c) 0cos1limsilim)cos1(si
limsicossi
lim00
20
20
!!
!
pppp x
x
x
x
x
xx
x
xxx
xxxx
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Example (Using 01liand01li !!gpgp xx xx
)
a)6
3
68
53lim
68
53lim !
!
5p5p
x
xx
x
xx b) !
!
6p6p
x
xx
xx
xx 11
23
lim1
23lim
34
c) 052
14
li52
4li
2
2
3
2
!
!
g7g7
x
xx
x
xx
xx d) g!
!
gpgp
x
xx
x
xx 11
23lim1
23lim34
.
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Definition
A line cx! iscalled a verticalasymptoteofafunction )(xf if 89
)(xf or gp)(xf as
@ cx or p cx .
A line Ly ! iscalled a horizontelasymptoteofafunction )(xf if Lxf p)( as gpx or
gpx .
Theorem
A line cx! isa verticalasymptoteofarationalfuntionxq
xpxf !)( ,if cx isafoctorofthe
denominator )(xq and notacommonfactorfor )(xp and )(xq .
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Example Findthe ve ticalandhorizontalasymptotes i any orthe ollowin unctions.
a)13)(
!
xxxf b)
4
)(2
!
x
xxf
c)4
8)(2
3
!x
xxf d)1
1)(23
!xxx
xxf
Solution
a) 1!x is the verticalasymptote.
11
1
31
lim1
3lim !
!
A
pA
p
x
xxx
xx. 1!@ y is the horizontalastmptote
b) 224
)(2
!
!xx
x
x
xxf , 2!x and 2!x are the verticalasymptotes.
041
1
lim4
lim
2
2!
!
gpgp
x
x
x
x
xx. 0!@ y is the horizontalastmptote
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c)
22228)(
2
2
!
!
xx
xxx
x
xxf 2!x istheverticalasymptote.
g!
!
gpgp
2
2
2
3
11
8
lim4
8lim
x
xx
x
xxx
. )(xf@ has no horizontalastmptotes
d) 1)1(1
1)1(
1
1
1)(2223
!
!
!
xx
x
xxx
x
xxx
xxf
)(xf@ has no vertical astmptote
01111
11
li1
1li
32
32
23!
!
gpgp
xxx
xx
xxx
x
xx.
0!@ y isthe horizontalastmptote
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Example Fi d a)63
2lim
2
gp xx
x b)
63
2lim
2
BC xx
x
c)thehorizontaland verticalasymptotesi any for .62)(
2
!
x
xxf
Note : xx !2
for 0"x and xx !2
for .0x
Sol tion
a) 31
63
21lim
/63/2
lim63
2lim
2222
!
!
!
D
E
D
E
D
E
x
xxxxx
xx
xxx
b) 31
63
21
lim/63/2
lim63
2lim
2222
!
!
!
F
G
F
GFG
x
xxxxx
xx
xxx
c) From(a)an(b),31
!y and3
!y arethehorizontalasymptotesand 2!x isthe vertical
asymptote for .62
)(2
!x
xxf
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Definition
A function )(xf issaid to becontinuousat cx! ifthefollowing conditionsaresatisfied.
a) )(cf is defined.
b) )(lim xfcxp
exists
c) )()(lim cfxfcx
!p
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Example Giventhe graph of )(xf .
Determine whether )(xf continuousat cx ! for 6,4,1!c .
Solution
At 1!x , 3)1( !f and )(lim1
xfx p
doesnotexist.
)(xf@ is discontinuousat 1!x .
At 4!x , 5)4( !f and 3)(li4
!
p
xfx
.
)(xf@ is discontinuousat 4!x .
At 6!x , 3)6( !f and 3)(lim6
!
p
xf
x
.
)(xf@ iscontinuousat 6!x .
5
3
-1
-2
-3
6
4
y
x
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Exa ple iven
"!
e
e
!
5if221
5if5
52if133
23if4
3if2
13
)( 2
xx
x
xx
xx
xxx
xf .
Deter ine whether )(xf continuousat cx! for 5,2,!c .
Solution At 3!x , 1)( !f and g!!
pp 6213lim)(lim
33 xxxf
xx. )(lim
3xf
x H doesnotexist.
)(xf@ is discontinuousat 3!x .
At 2!x , 21)32(3)2( 2 !!f , 264lim)(lim22
!!
ppxxf
xx
,
21)3(3lim)(lim 222
!!
II
xxfxx
2)(li2
!p
xfx
. )(xf@ is continuousat 2!x .
At 5!x , 5)5( !f ,
111)3(3lim)(lim 255
!!
PP
xxfxx
, .11221lim)(lim55
!!
pp
xxfxx
)5(11)(li2
fxfx
{!
p. )(xf@ is discontinuousat 5!x .
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Derivative of a function
romthe diagramx
Gradiaenth
xfhxfAB
)()( !
Gradiaentofthe curve ath
xfhxfA
h
)()(li
0
!
p
If ),(xfy ! then
.)()(
li)(
d
d
0
,
h
xfhxfxf
x
y
h
!!
Q
This process is known asdifferentiation fromthe first principlesor differentiation using definition.
)(xf
y
x
)( hxf
x x
B
A
)()( xfhxf
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Example Findthederivativeof2
1)(x
xf ! withrespecttoxfromthefirstprinciples.
Solutionh
xfhxfxf
x
y
h
)()(lim)(
0
, !!p
34
220
22
2
0
22
222
0
22
22
0
22
0
22
2li
2li
2li
li
11
li
xx
x
xhx
hx
xhxh
hxh
xhxh
hxhxx
xhxh
hxx
h
xhx
h
h
h
h
h
!!
!
!
!
!
!
p
p
p
p
p
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Example
Given 22)( 2 ! xxxf .Find )(' xf fromthe firstprinciples.
Solutionh
xfhxfxf
h
)()(lim)(
0
, !
R
14
124lim
24lim
222242lim
2222lim
0
2
0
222
0
22
0
!
!
!
!
!
S
S
S
S
x
x
x
xxxxx
xxxx
T
T
T
T
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Example
Given 21
)( xxf ! .Find )(' xf fromthefirstprinciples.
Solutionh
xfhxfxf
h
)()(lim)(
0
, !
U
x
xhx
xhxh
h
xhxh xh
x
xhx
xhxh
xhx
h
h
h
h
2
1
1li
li
li
li
0
0
0
0
!
!
!
!
v
!
p
p
p
p
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E a ple Show that .sis xxx
!
Solution )('cosdd
xfxx ! for .cos)( xxf !
Fro the first principles
h
xfhxfxf
h
)()(lim)(
0
, !
V
xxx
hx
hx
hx
h
xh
x
h
x
h
xx
hxxx
h
xhx
hh
hh
h
h
h
h
sin)1(sin)(c s
sinhlimsin1c sh
limc s
sinhsinlim1c shc s
lim
sinhsin1c shc slim
sinhsin1c shc slim
c ssinhsinc shc slim
c s)c s(
lim
!!
!
!
!
!
!
!
pp
pp
p
p
p
p
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