Lesson 4: Calculating Limits
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Transcript of Lesson 4: Calculating Limits
. . . . . .
Section1.4CalculatingLimits
V63.0121.034, CalculusI
September14, 2009
Announcements
I FirstwrittenassignmentdueWednesdayI FirstwebassignmentdueMonday, September21
. . . . . .
Outline
LimitsandPathologies
BasicLimits
LimitLawsThedirectsubstitutionproperty
LimitswithAlgebraTwomorelimittheorems
Twoimportanttrigonometriclimits
. . . . . .
HeuristicDefinitionofaLimit
DefinitionWewrite
limx→a
f(x) = L
andsay
“thelimitof f(x), as x approaches a, equals L”
ifwecanmakethevaluesof f(x) arbitrarilycloseto L (asclosetoL aswelike)bytaking x tobesufficientlycloseto a (oneithersideof a)butnotequalto a.
. . . . . .
Theerror-tolerancegame
A gamebetweentwoplayerstodecideifalimit limx→a
f(x) exists.
I Player1: Choose L tobethelimit.I Player2: Proposean“error”levelaround L.I Player1: Choosea“tolerance”levelaround a sothat
x-pointswithinthattolerancelevelaretakento y-valueswithintheerrorlevel.
IfPlayer1canalwayswin, limx→a
f(x) = L.
. . . . . .
ExampleFind lim
x→0x2 ifitexists.
Solution
I I claimthelimitiszero.I Iftheerrorlevelis 0.01, I needtoguaranteethat
−0.01 < x2 < 0.01 forall x sufficientlyclosetozero.I If −0.1 < x < 0.1, then 0 ≤ x2 < 0.01, soI winthatround.I Whatshouldthetolerancebeiftheerroris 0.0001?
Bysettingtoleranceequaltothesquarerootoftheerror, wecanguaranteetobewithinanyerror.
. . . . . .
ExampleFind lim
x→0x2 ifitexists.
Solution
I I claimthelimitiszero.
I Iftheerrorlevelis 0.01, I needtoguaranteethat−0.01 < x2 < 0.01 forall x sufficientlyclosetozero.
I If −0.1 < x < 0.1, then 0 ≤ x2 < 0.01, soI winthatround.I Whatshouldthetolerancebeiftheerroris 0.0001?
Bysettingtoleranceequaltothesquarerootoftheerror, wecanguaranteetobewithinanyerror.
. . . . . .
ExampleFind lim
x→0x2 ifitexists.
Solution
I I claimthelimitiszero.I Iftheerrorlevelis 0.01, I needtoguaranteethat
−0.01 < x2 < 0.01 forall x sufficientlyclosetozero.
I If −0.1 < x < 0.1, then 0 ≤ x2 < 0.01, soI winthatround.I Whatshouldthetolerancebeiftheerroris 0.0001?
Bysettingtoleranceequaltothesquarerootoftheerror, wecanguaranteetobewithinanyerror.
. . . . . .
ExampleFind lim
x→0x2 ifitexists.
Solution
I I claimthelimitiszero.I Iftheerrorlevelis 0.01, I needtoguaranteethat
−0.01 < x2 < 0.01 forall x sufficientlyclosetozero.I If −0.1 < x < 0.1, then 0 ≤ x2 < 0.01, soI winthatround.
I Whatshouldthetolerancebeiftheerroris 0.0001?
Bysettingtoleranceequaltothesquarerootoftheerror, wecanguaranteetobewithinanyerror.
. . . . . .
ExampleFind lim
x→0x2 ifitexists.
Solution
I I claimthelimitiszero.I Iftheerrorlevelis 0.01, I needtoguaranteethat
−0.01 < x2 < 0.01 forall x sufficientlyclosetozero.I If −0.1 < x < 0.1, then 0 ≤ x2 < 0.01, soI winthatround.I Whatshouldthetolerancebeiftheerroris 0.0001?
Bysettingtoleranceequaltothesquarerootoftheerror, wecanguaranteetobewithinanyerror.
. . . . . .
ExampleFind lim
x→0x2 ifitexists.
Solution
I I claimthelimitiszero.I Iftheerrorlevelis 0.01, I needtoguaranteethat
−0.01 < x2 < 0.01 forall x sufficientlyclosetozero.I If −0.1 < x < 0.1, then 0 ≤ x2 < 0.01, soI winthatround.I Whatshouldthetolerancebeiftheerroris 0.0001?
Bysettingtoleranceequaltothesquarerootoftheerror, wecanguaranteetobewithinanyerror.
. . . . . .
Example
Find limx→0
|x|x
ifitexists.
Solution
Thefunctioncanalsobewrittenas
|x|x
=
{1 if x > 0;
−1 if x < 0
Whatwouldbethelimit?Theerror-tolerancegamefails, but
limx→0+
f(x) = 1 limx→0−
f(x) = −1
. . . . . .
Example
Find limx→0
|x|x
ifitexists.
SolutionThefunctioncanalsobewrittenas
|x|x
=
{1 if x > 0;
−1 if x < 0
Whatwouldbethelimit?
Theerror-tolerancegamefails, but
limx→0+
f(x) = 1 limx→0−
f(x) = −1
. . . . . .
Theerror-tolerancegame
. .x
.y
..−1
..1 .
.
.Part of graph in-side blue is notinside green
.Part of graph in-side blue is notinside green
I Thesearetheonlygoodchoices; thelimitdoesnotexist.
. . . . . .
Theerror-tolerancegame
. .x
.y
..−1
..1 .
.
.Part of graph in-side blue is notinside green
.Part of graph in-side blue is notinside green
I Thesearetheonlygoodchoices; thelimitdoesnotexist.
. . . . . .
Theerror-tolerancegame
. .x
.y
..−1
..1 .
.
.Part of graph in-side blue is notinside green
.Part of graph in-side blue is notinside green
I Thesearetheonlygoodchoices; thelimitdoesnotexist.
. . . . . .
Theerror-tolerancegame
. .x
.y
..−1
..1 .
.
.Part of graph in-side blue is notinside green
.Part of graph in-side blue is notinside green
I Thesearetheonlygoodchoices; thelimitdoesnotexist.
. . . . . .
Theerror-tolerancegame
. .x
.y
..−1
..1 .
.
.Part of graph in-side blue is notinside green
.Part of graph in-side blue is notinside green
I Thesearetheonlygoodchoices; thelimitdoesnotexist.
. . . . . .
Theerror-tolerancegame
. .x
.y
..−1
..1 .
.
.Part of graph in-side blue is notinside green
.Part of graph in-side blue is notinside green
I Thesearetheonlygoodchoices; thelimitdoesnotexist.
. . . . . .
Theerror-tolerancegame
. .x
.y
..−1
..1 .
.
.Part of graph in-side blue is notinside green
.Part of graph in-side blue is notinside green
I Thesearetheonlygoodchoices; thelimitdoesnotexist.
. . . . . .
Theerror-tolerancegame
. .x
.y
..−1
..1 .
.
.Part of graph in-side blue is notinside green
.Part of graph in-side blue is notinside green
I Thesearetheonlygoodchoices; thelimitdoesnotexist.
. . . . . .
Theerror-tolerancegame
. .x
.y
..−1
..1 .
.
.Part of graph in-side blue is notinside green
.Part of graph in-side blue is notinside green
I Thesearetheonlygoodchoices; thelimitdoesnotexist.
. . . . . .
One-sidedlimits
DefinitionWewrite
limx→a+
f(x) = L
andsay
“thelimitof f(x), as x approaches a fromthe right, equals L”
ifwecanmakethevaluesof f(x) arbitrarilycloseto L (asclosetoL aswelike)bytaking x tobesufficientlycloseto a (oneithersideof a)and greater than a.
. . . . . .
One-sidedlimits
DefinitionWewrite
limx→a−
f(x) = L
andsay
“thelimitof f(x), as x approaches a fromthe left, equals L”
ifwecanmakethevaluesof f(x) arbitrarilycloseto L (asclosetoL aswelike)bytaking x tobesufficientlycloseto a (oneithersideof a)and less than a.
. . . . . .
Example
Find limx→0
|x|x
ifitexists.
SolutionThefunctioncanalsobewrittenas
|x|x
=
{1 if x > 0;
−1 if x < 0
Whatwouldbethelimit?Theerror-tolerancegamefails, but
limx→0+
f(x) = 1 limx→0−
f(x) = −1
. . . . . .
Example
Find limx→0+
1xifitexists.
SolutionThelimitdoesnotexistbecausethefunctionisunboundednear0. Nextweekwewillunderstandthestatementthat
limx→0+
1x
= +∞
. . . . . .
Theerror-tolerancegame
. .x
.y
.0
..L?
.The graph escapes thegreen, so no good.Evenworse!
.The limit does not existbecause the function isunbounded near 0
. . . . . .
Theerror-tolerancegame
. .x
.y
.0
..L?
.The graph escapes thegreen, so no good.Evenworse!
.The limit does not existbecause the function isunbounded near 0
. . . . . .
Theerror-tolerancegame
. .x
.y
.0
..L?
.The graph escapes thegreen, so no good.Evenworse!
.The limit does not existbecause the function isunbounded near 0
. . . . . .
Theerror-tolerancegame
. .x
.y
.0
..L?
.The graph escapes thegreen, so no good
.Evenworse!
.The limit does not existbecause the function isunbounded near 0
. . . . . .
Theerror-tolerancegame
. .x
.y
.0
..L?
.The graph escapes thegreen, so no good.Evenworse!
.The limit does not existbecause the function isunbounded near 0
. . . . . .
Theerror-tolerancegame
. .x
.y
.0
..L?
.The graph escapes thegreen, so no good
.Evenworse!
.The limit does not existbecause the function isunbounded near 0
. . . . . .
Theerror-tolerancegame
. .x
.y
.0
..L?
.The graph escapes thegreen, so no good.Evenworse!
.The limit does not existbecause the function isunbounded near 0
. . . . . .
Example
Find limx→0+
1xifitexists.
SolutionThelimitdoesnotexistbecausethefunctionisunboundednear0. Nextweekwewillunderstandthestatementthat
limx→0+
1x
= +∞
. . . . . .
Weird, wildstuff
ExampleFind lim
x→0sin
(π
x
)ifitexists.
I f(x) = 0 when x =
1kforanyinteger k
I f(x) = 1 when x =
12k + 1/2
foranyinteger k
I f(x) = −1 when x =
12k− 1/2
foranyinteger k
. . . . . .
Weird, wildstuff
ExampleFind lim
x→0sin
(π
x
)ifitexists.
I f(x) = 0 when x =
1kforanyinteger k
I f(x) = 1 when x =
12k + 1/2
foranyinteger k
I f(x) = −1 when x =
12k− 1/2
foranyinteger k
. . . . . .
Weird, wildstuff
ExampleFind lim
x→0sin
(π
x
)ifitexists.
I f(x) = 0 when x =1kforanyinteger k
I f(x) = 1 when x =
12k + 1/2
foranyinteger k
I f(x) = −1 when x =
12k− 1/2
foranyinteger k
. . . . . .
Weird, wildstuff
ExampleFind lim
x→0sin
(π
x
)ifitexists.
I f(x) = 0 when x =1kforanyinteger k
I f(x) = 1 when x =1
2k + 1/2foranyinteger k
I f(x) = −1 when x =
12k− 1/2
foranyinteger k
. . . . . .
Weird, wildstuff
ExampleFind lim
x→0sin
(π
x
)ifitexists.
I f(x) = 0 when x =1kforanyinteger k
I f(x) = 1 when x =1
2k + 1/2foranyinteger k
I f(x) = −1 when x =1
2k− 1/2foranyinteger k
. . . . . .
Weird, wildstuffcontinued
Hereisagraphofthefunction:
. .x
.y
..−1
..1
Thereareinfinitelymanypointsarbitrarilyclosetozerowheref(x) is 0, or 1, or −1. Sothelimitcannotexist.
. . . . . .
Whatcouldgowrong?SummaryofLimitPathologies
Howcouldafunctionfailtohavealimit? Somepossibilities:I left-andright-handlimitsexistbutarenotequalI Thefunctionisunboundednear a (possibleinfinitelimits,morelater)
I Oscillationwithincreasinglyhighfrequencynear a
. . . . . .
MeettheMathematician: AugustinLouisCauchy
I French, 1789–1857I RoyalistandCatholicI madecontributionsingeometry, calculus,complexanalysis,numbertheory
I createdthedefinitionoflimitweusetodaybutdidn’tunderstandit
. . . . . .
Outline
LimitsandPathologies
BasicLimits
LimitLawsThedirectsubstitutionproperty
LimitswithAlgebraTwomorelimittheorems
Twoimportanttrigonometriclimits
. . . . . .
Reallybasiclimits
FactLet c beaconstantand a arealnumber.
(i) limx→a
x = a
(ii) limx→a
c = c
Proof.Thefirstistautological, thesecondistrivial.
. . . . . .
Reallybasiclimits
FactLet c beaconstantand a arealnumber.
(i) limx→a
x = a
(ii) limx→a
c = c
Proof.Thefirstistautological, thesecondistrivial.
. . . . . .
ET gamefor f(x) = x
. .x
.y
..a
..a
I Settingerrorequaltotoleranceworks!
. . . . . .
ET gamefor f(x) = x
. .x
.y
..a
..a
I Settingerrorequaltotoleranceworks!
. . . . . .
ET gamefor f(x) = x
. .x
.y
..a
..a
I Settingerrorequaltotoleranceworks!
. . . . . .
ET gamefor f(x) = x
. .x
.y
..a
..a
I Settingerrorequaltotoleranceworks!
. . . . . .
ET gamefor f(x) = x
. .x
.y
..a
..a
I Settingerrorequaltotoleranceworks!
. . . . . .
ET gamefor f(x) = x
. .x
.y
..a
..a
I Settingerrorequaltotoleranceworks!
. . . . . .
ET gamefor f(x) = x
. .x
.y
..a
..a
I Settingerrorequaltotoleranceworks!
. . . . . .
ET gamefor f(x) = c
.
.x
.y
..a
..c
I anytoleranceworks!
. . . . . .
ET gamefor f(x) = c
. .x
.y
..a
..c
I anytoleranceworks!
. . . . . .
ET gamefor f(x) = c
. .x
.y
..a
..c
I anytoleranceworks!
. . . . . .
ET gamefor f(x) = c
. .x
.y
..a
..c
I anytoleranceworks!
. . . . . .
ET gamefor f(x) = c
. .x
.y
..a
..c
I anytoleranceworks!
. . . . . .
ET gamefor f(x) = c
. .x
.y
..a
..c
I anytoleranceworks!
. . . . . .
ET gamefor f(x) = c
. .x
.y
..a
..c
I anytoleranceworks!
. . . . . .
Reallybasiclimits
FactLet c beaconstantand a arealnumber.
(i) limx→a
x = a
(ii) limx→a
c = c
Proof.Thefirstistautological, thesecondistrivial.
. . . . . .
Outline
LimitsandPathologies
BasicLimits
LimitLawsThedirectsubstitutionproperty
LimitswithAlgebraTwomorelimittheorems
Twoimportanttrigonometriclimits
. . . . . .
Limitsandarithmetic
FactSuppose lim
x→af(x) and lim
x→ag(x) existand c isaconstant. Then
1. limx→a
[f(x) + g(x)] = limx→a
f(x) + limx→a
g(x)
(errorsadd)
2. limx→a
[f(x) − g(x)] = limx→a
f(x) − limx→a
g(x)
(combinationofadding
andscaling)
3. limx→a
[cf(x)] = c limx→a
f(x)
(errorscales)
4. limx→a
[f(x)g(x)] = limx→a
f(x) · limx→a
g(x)
(morecomplicated, but
doable)
. . . . . .
Limitsandarithmetic
FactSuppose lim
x→af(x) and lim
x→ag(x) existand c isaconstant. Then
1. limx→a
[f(x) + g(x)] = limx→a
f(x) + limx→a
g(x) (errorsadd)
2. limx→a
[f(x) − g(x)] = limx→a
f(x) − limx→a
g(x)
(combinationofadding
andscaling)
3. limx→a
[cf(x)] = c limx→a
f(x)
(errorscales)
4. limx→a
[f(x)g(x)] = limx→a
f(x) · limx→a
g(x)
(morecomplicated, but
doable)
. . . . . .
Limitsandarithmetic
FactSuppose lim
x→af(x) and lim
x→ag(x) existand c isaconstant. Then
1. limx→a
[f(x) + g(x)] = limx→a
f(x) + limx→a
g(x) (errorsadd)
2. limx→a
[f(x) − g(x)] = limx→a
f(x) − limx→a
g(x)
(combinationofadding
andscaling)
3. limx→a
[cf(x)] = c limx→a
f(x)
(errorscales)
4. limx→a
[f(x)g(x)] = limx→a
f(x) · limx→a
g(x)
(morecomplicated, but
doable)
. . . . . .
Limitsandarithmetic
FactSuppose lim
x→af(x) and lim
x→ag(x) existand c isaconstant. Then
1. limx→a
[f(x) + g(x)] = limx→a
f(x) + limx→a
g(x) (errorsadd)
2. limx→a
[f(x) − g(x)] = limx→a
f(x) − limx→a
g(x)
(combinationofadding
andscaling)
3. limx→a
[cf(x)] = c limx→a
f(x)
(errorscales)
4. limx→a
[f(x)g(x)] = limx→a
f(x) · limx→a
g(x)
(morecomplicated, but
doable)
. . . . . .
Limitsandarithmetic
FactSuppose lim
x→af(x) and lim
x→ag(x) existand c isaconstant. Then
1. limx→a
[f(x) + g(x)] = limx→a
f(x) + limx→a
g(x) (errorsadd)
2. limx→a
[f(x) − g(x)] = limx→a
f(x) − limx→a
g(x)
(combinationofadding
andscaling)
3. limx→a
[cf(x)] = c limx→a
f(x) (errorscales)
4. limx→a
[f(x)g(x)] = limx→a
f(x) · limx→a
g(x)
(morecomplicated, but
doable)
. . . . . .
Limitsandarithmetic
FactSuppose lim
x→af(x) and lim
x→ag(x) existand c isaconstant. Then
1. limx→a
[f(x) + g(x)] = limx→a
f(x) + limx→a
g(x) (errorsadd)
2. limx→a
[f(x) − g(x)] = limx→a
f(x) − limx→a
g(x) (combinationofadding
andscaling)
3. limx→a
[cf(x)] = c limx→a
f(x) (errorscales)
4. limx→a
[f(x)g(x)] = limx→a
f(x) · limx→a
g(x)
(morecomplicated, but
doable)
. . . . . .
Limitsandarithmetic
FactSuppose lim
x→af(x) and lim
x→ag(x) existand c isaconstant. Then
1. limx→a
[f(x) + g(x)] = limx→a
f(x) + limx→a
g(x) (errorsadd)
2. limx→a
[f(x) − g(x)] = limx→a
f(x) − limx→a
g(x) (combinationofadding
andscaling)
3. limx→a
[cf(x)] = c limx→a
f(x) (errorscales)
4. limx→a
[f(x)g(x)] = limx→a
f(x) · limx→a
g(x)
(morecomplicated, but
doable)
. . . . . .
Limitsandarithmetic
FactSuppose lim
x→af(x) and lim
x→ag(x) existand c isaconstant. Then
1. limx→a
[f(x) + g(x)] = limx→a
f(x) + limx→a
g(x) (errorsadd)
2. limx→a
[f(x) − g(x)] = limx→a
f(x) − limx→a
g(x) (combinationofadding
andscaling)
3. limx→a
[cf(x)] = c limx→a
f(x) (errorscales)
4. limx→a
[f(x)g(x)] = limx→a
f(x) · limx→a
g(x) (morecomplicated, but
doable)
. . . . . .
LimitsandarithmeticII
Fact(Continued)
5. limx→a
f(x)g(x)
=limx→a
f(x)
limx→a
g(x), if lim
x→ag(x) ̸= 0.
6. limx→a
[f(x)]n =[limx→a
f(x)]n
(followsfrom4repeatedly)
7. limx→a
xn = an
(followsfrom6)
8. limx→a
n√x = n
√a
9. limx→a
n√
f(x) = n
√limx→a
f(x) (If n iseven, wemustadditionally
assumethat limx→a
f(x) > 0)
. . . . . .
LimitsandarithmeticII
Fact(Continued)
5. limx→a
f(x)g(x)
=limx→a
f(x)
limx→a
g(x), if lim
x→ag(x) ̸= 0.
6. limx→a
[f(x)]n =[limx→a
f(x)]n
(followsfrom4repeatedly)
7. limx→a
xn = an
(followsfrom6)
8. limx→a
n√x = n
√a
9. limx→a
n√
f(x) = n
√limx→a
f(x) (If n iseven, wemustadditionally
assumethat limx→a
f(x) > 0)
. . . . . .
LimitsandarithmeticII
Fact(Continued)
5. limx→a
f(x)g(x)
=limx→a
f(x)
limx→a
g(x), if lim
x→ag(x) ̸= 0.
6. limx→a
[f(x)]n =[limx→a
f(x)]n
(followsfrom4repeatedly)
7. limx→a
xn = an
(followsfrom6)
8. limx→a
n√x = n
√a
9. limx→a
n√
f(x) = n
√limx→a
f(x) (If n iseven, wemustadditionally
assumethat limx→a
f(x) > 0)
. . . . . .
LimitsandarithmeticII
Fact(Continued)
5. limx→a
f(x)g(x)
=limx→a
f(x)
limx→a
g(x), if lim
x→ag(x) ̸= 0.
6. limx→a
[f(x)]n =[limx→a
f(x)]n
(followsfrom4repeatedly)
7. limx→a
xn = an
(followsfrom6)
8. limx→a
n√x = n
√a
9. limx→a
n√
f(x) = n
√limx→a
f(x) (If n iseven, wemustadditionally
assumethat limx→a
f(x) > 0)
. . . . . .
LimitsandarithmeticII
Fact(Continued)
5. limx→a
f(x)g(x)
=limx→a
f(x)
limx→a
g(x), if lim
x→ag(x) ̸= 0.
6. limx→a
[f(x)]n =[limx→a
f(x)]n
(followsfrom4repeatedly)
7. limx→a
xn = an
(followsfrom6)
8. limx→a
n√x = n
√a
9. limx→a
n√
f(x) = n
√limx→a
f(x) (If n iseven, wemustadditionally
assumethat limx→a
f(x) > 0)
. . . . . .
LimitsandarithmeticII
Fact(Continued)
5. limx→a
f(x)g(x)
=limx→a
f(x)
limx→a
g(x), if lim
x→ag(x) ̸= 0.
6. limx→a
[f(x)]n =[limx→a
f(x)]n
(followsfrom4repeatedly)
7. limx→a
xn = an (followsfrom6)
8. limx→a
n√x = n
√a
9. limx→a
n√
f(x) = n
√limx→a
f(x) (If n iseven, wemustadditionally
assumethat limx→a
f(x) > 0)
. . . . . .
LimitsandarithmeticII
Fact(Continued)
5. limx→a
f(x)g(x)
=limx→a
f(x)
limx→a
g(x), if lim
x→ag(x) ̸= 0.
6. limx→a
[f(x)]n =[limx→a
f(x)]n
(followsfrom4repeatedly)
7. limx→a
xn = an (followsfrom6)
8. limx→a
n√x = n
√a
9. limx→a
n√
f(x) = n
√limx→a
f(x) (If n iseven, wemustadditionally
assumethat limx→a
f(x) > 0)
. . . . . .
Applyingthelimitlaws
ExampleFind lim
x→3
(x2 + 2x + 4
).
SolutionByapplyingthelimitlawsrepeatedly:
limx→3
(x2 + 2x + 4
)
= limx→3
(x2
)+ lim
x→3(2x) + lim
x→3(4)
=
(limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
. . . . . .
Applyingthelimitlaws
ExampleFind lim
x→3
(x2 + 2x + 4
).
SolutionByapplyingthelimitlawsrepeatedly:
limx→3
(x2 + 2x + 4
)
= limx→3
(x2
)+ lim
x→3(2x) + lim
x→3(4)
=
(limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
. . . . . .
Applyingthelimitlaws
ExampleFind lim
x→3
(x2 + 2x + 4
).
SolutionByapplyingthelimitlawsrepeatedly:
limx→3
(x2 + 2x + 4
)= lim
x→3
(x2
)+ lim
x→3(2x) + lim
x→3(4)
=
(limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
. . . . . .
Applyingthelimitlaws
ExampleFind lim
x→3
(x2 + 2x + 4
).
SolutionByapplyingthelimitlawsrepeatedly:
limx→3
(x2 + 2x + 4
)= lim
x→3
(x2
)+ lim
x→3(2x) + lim
x→3(4)
=
(limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
. . . . . .
Applyingthelimitlaws
ExampleFind lim
x→3
(x2 + 2x + 4
).
SolutionByapplyingthelimitlawsrepeatedly:
limx→3
(x2 + 2x + 4
)= lim
x→3
(x2
)+ lim
x→3(2x) + lim
x→3(4)
=
(limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
. . . . . .
Applyingthelimitlaws
ExampleFind lim
x→3
(x2 + 2x + 4
).
SolutionByapplyingthelimitlawsrepeatedly:
limx→3
(x2 + 2x + 4
)= lim
x→3
(x2
)+ lim
x→3(2x) + lim
x→3(4)
=
(limx→3
x)2
+ 2 · limx→3
(x) + 4
= (3)2 + 2 · 3 + 4
= 9 + 6 + 4 = 19.
. . . . . .
Yourturn
Example
Find limx→3
x2 + 2x + 4x3 + 11
SolutionTheansweris
1938
=12.
. . . . . .
Yourturn
Example
Find limx→3
x2 + 2x + 4x3 + 11
SolutionTheansweris
1938
=12.
. . . . . .
DirectSubstitutionProperty
Theorem(TheDirectSubstitutionProperty)If f isapolynomialorarationalfunctionand a isinthedomainoff, then
limx→a
f(x) = f(a)
. . . . . .
Outline
LimitsandPathologies
BasicLimits
LimitLawsThedirectsubstitutionproperty
LimitswithAlgebraTwomorelimittheorems
Twoimportanttrigonometriclimits
. . . . . .
Limitsdonotseethepoint! (inagoodway)
TheoremIf f(x) = g(x) when x ̸= a, and lim
x→ag(x) = L, then lim
x→af(x) = L.
Example
Find limx→−1
x2 + 2x + 1x + 1
, ifitexists.
SolutionSince
x2 + 2x + 1x + 1
= x + 1 whenever x ̸= −1, andsince
limx→−1
x + 1 = 0, wehave limx→−1
x2 + 2x + 1x + 1
= 0.
. . . . . .
Limitsdonotseethepoint! (inagoodway)
TheoremIf f(x) = g(x) when x ̸= a, and lim
x→ag(x) = L, then lim
x→af(x) = L.
Example
Find limx→−1
x2 + 2x + 1x + 1
, ifitexists.
SolutionSince
x2 + 2x + 1x + 1
= x + 1 whenever x ̸= −1, andsince
limx→−1
x + 1 = 0, wehave limx→−1
x2 + 2x + 1x + 1
= 0.
. . . . . .
Limitsdonotseethepoint! (inagoodway)
TheoremIf f(x) = g(x) when x ̸= a, and lim
x→ag(x) = L, then lim
x→af(x) = L.
Example
Find limx→−1
x2 + 2x + 1x + 1
, ifitexists.
SolutionSince
x2 + 2x + 1x + 1
= x + 1 whenever x ̸= −1, andsince
limx→−1
x + 1 = 0, wehave limx→−1
x2 + 2x + 1x + 1
= 0.
. . . . . .
ET gamefor f(x) =x2 + 2x + 1
x + 1
. .x
.y
...−1
I Evenif f(−1) weresomethingelse, itwouldnoteffectthelimit.
. . . . . .
ET gamefor f(x) =x2 + 2x + 1
x + 1
. .x
.y
...−1
I Evenif f(−1) weresomethingelse, itwouldnoteffectthelimit.
. . . . . .
Limitofafunctiondefinedpiecewiseataboundarypoint
ExampleLet
f(x) =
{x2 x ≥ 0
−x x < 0
Does limx→0
f(x) exist?
.
SolutionWehave
limx→0+
f(x) MTP= lim
x→0+x2 DSP
= 02 = 0
Likewise:limx→0−
f(x) = limx→0−
−x = −0 = 0
So limx→0
f(x) = 0.
. . . . . .
Limitofafunctiondefinedpiecewiseataboundarypoint
ExampleLet
f(x) =
{x2 x ≥ 0
−x x < 0
Does limx→0
f(x) exist?
.
SolutionWehave
limx→0+
f(x) MTP= lim
x→0+x2 DSP
= 02 = 0
Likewise:limx→0−
f(x) = limx→0−
−x = −0 = 0
So limx→0
f(x) = 0.
. . . . . .
Limitofafunctiondefinedpiecewiseataboundarypoint
ExampleLet
f(x) =
{x2 x ≥ 0
−x x < 0
Does limx→0
f(x) exist?
.
SolutionWehave
limx→0+
f(x) MTP= lim
x→0+x2 DSP
= 02 = 0
Likewise:limx→0−
f(x) = limx→0−
−x = −0 = 0
So limx→0
f(x) = 0.
. . . . . .
Findinglimitsbyalgebraicmanipulations
Example
Find limx→4
√x− 2x− 4
.
SolutionWritethedenominatoras x− 4 =
√x2 − 4 = (
√x− 2)(
√x + 2).
So
limx→4
√x− 2x− 4
= limx→4
√x− 2
(√x− 2)(
√x + 2)
= limx→4
1√x + 2
=14
. . . . . .
Findinglimitsbyalgebraicmanipulations
Example
Find limx→4
√x− 2x− 4
.
SolutionWritethedenominatoras x− 4 =
√x2 − 4 = (
√x− 2)(
√x + 2).
So
limx→4
√x− 2x− 4
= limx→4
√x− 2
(√x− 2)(
√x + 2)
= limx→4
1√x + 2
=14
. . . . . .
Findinglimitsbyalgebraicmanipulations
Example
Find limx→4
√x− 2x− 4
.
SolutionWritethedenominatoras x− 4 =
√x2 − 4 = (
√x− 2)(
√x + 2).
So
limx→4
√x− 2x− 4
= limx→4
√x− 2
(√x− 2)(
√x + 2)
= limx→4
1√x + 2
=14
. . . . . .
Yourturn
ExampleLet
f(x) =
{1− x2 x ≥ 1
2x x < 1
Find limx→1
f(x) ifitexists.
. ..1
.
.
SolutionWehave
limx→1+
f(x) = limx→1+
(1− x2
) DSP= 0
limx→1−
f(x) = limx→1−
(2x) DSP= 2
Theleft-andright-handlimitsdisagree, sothelimitdoesnotexist.
. . . . . .
Yourturn
ExampleLet
f(x) =
{1− x2 x ≥ 1
2x x < 1
Find limx→1
f(x) ifitexists.
. ..1
.
.
SolutionWehave
limx→1+
f(x) = limx→1+
(1− x2
) DSP= 0
limx→1−
f(x) = limx→1−
(2x) DSP= 2
Theleft-andright-handlimitsdisagree, sothelimitdoesnotexist.
. . . . . .
Yourturn
ExampleLet
f(x) =
{1− x2 x ≥ 1
2x x < 1
Find limx→1
f(x) ifitexists.
. ..1
.
.
SolutionWehave
limx→1+
f(x) = limx→1+
(1− x2
) DSP= 0
limx→1−
f(x) = limx→1−
(2x) DSP= 2
Theleft-andright-handlimitsdisagree, sothelimitdoesnotexist.
. . . . . .
TwoMoreImportantLimitTheorems
TheoremIf f(x) ≤ g(x) when x isnear a (exceptpossiblyat a), then
limx→a
f(x) ≤ limx→a
g(x)
(asusual, providedtheselimitsexist).
Theorem(TheSqueeze/Sandwich/PinchingTheorem)If f(x) ≤ g(x) ≤ h(x) when x isnear a (asusual, exceptpossiblyata), and
limx→a
f(x) = limx→a
h(x) = L,
thenlimx→a
g(x) = L.
. . . . . .
WecanusetheSqueezeTheoremtomakecomplicatedlimitssimple.
ExampleShowthat lim
x→0x2 sin
(π
x
)= 0.
SolutionWehaveforall x,
−x2 ≤ x2 sin(π
x
)≤ x2
Theleftandrightsidesgotozeroas x → 0.
. . . . . .
WecanusetheSqueezeTheoremtomakecomplicatedlimitssimple.
ExampleShowthat lim
x→0x2 sin
(π
x
)= 0.
SolutionWehaveforall x,
−x2 ≤ x2 sin(π
x
)≤ x2
Theleftandrightsidesgotozeroas x → 0.
. . . . . .
WecanusetheSqueezeTheoremtomakecomplicatedlimitssimple.
ExampleShowthat lim
x→0x2 sin
(π
x
)= 0.
SolutionWehaveforall x,
−x2 ≤ x2 sin(π
x
)≤ x2
Theleftandrightsidesgotozeroas x → 0.
. . . . . .
IllustrationoftheSqueezeTheorem
. .x
.y .h(x) = x2
.f(x) = −x2
.g(x) = x2 sin(1x
)
. . . . . .
IllustrationoftheSqueezeTheorem
. .x
.y .h(x) = x2
.f(x) = −x2
.g(x) = x2 sin(1x
)
. . . . . .
IllustrationoftheSqueezeTheorem
. .x
.y .h(x) = x2
.f(x) = −x2
.g(x) = x2 sin(1x
)
. . . . . .
Outline
LimitsandPathologies
BasicLimits
LimitLawsThedirectsubstitutionproperty
LimitswithAlgebraTwomorelimittheorems
Twoimportanttrigonometriclimits
. . . . . .
Twoimportanttrigonometriclimits
TheoremThefollowingtwolimitshold:
I limθ→0
sin θ
θ= 1
I limθ→0
cos θ − 1θ
= 0
. . . . . .
ProofoftheSineLimit
Proof.
. .θ
.sin θ
.cos θ
.θ
.tan θ
.−1 .1
Notice
sin θ ≤
θ
≤ 2 tanθ
2≤ tan θ
Divideby sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Takereciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, theleftandrightsidestendto 1. So, then, mustthemiddleexpression.
. . . . . .
ProofoftheSineLimit
Proof.
. .θ.sin θ
.cos θ
.θ
.tan θ
.−1 .1
Notice
sin θ ≤ θ
≤ 2 tanθ
2≤ tan θ
Divideby sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Takereciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, theleftandrightsidestendto 1. So, then, mustthemiddleexpression.
. . . . . .
ProofoftheSineLimit
Proof.
. .θ.sin θ
.cos θ
.θ .tan θ
.−1 .1
Notice
sin θ ≤ θ
≤ 2 tanθ
2≤
tan θ
Divideby sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Takereciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, theleftandrightsidestendto 1. So, then, mustthemiddleexpression.
. . . . . .
ProofoftheSineLimit
Proof.
. .θ.sin θ
.cos θ
.θ .tan θ
.−1 .1
Notice
sin θ ≤ θ ≤ 2 tanθ
2≤ tan θ
Divideby sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Takereciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, theleftandrightsidestendto 1. So, then, mustthemiddleexpression.
. . . . . .
ProofoftheSineLimit
Proof.
. .θ.sin θ
.cos θ
.θ .tan θ
.−1 .1
Notice
sin θ ≤ θ ≤ 2 tanθ
2≤ tan θ
Divideby sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Takereciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, theleftandrightsidestendto 1. So, then, mustthemiddleexpression.
. . . . . .
ProofoftheSineLimit
Proof.
. .θ.sin θ
.cos θ
.θ .tan θ
.−1 .1
Notice
sin θ ≤ θ ≤ 2 tanθ
2≤ tan θ
Divideby sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Takereciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, theleftandrightsidestendto 1. So, then, mustthemiddleexpression.
. . . . . .
ProofoftheSineLimit
Proof.
. .θ.sin θ
.cos θ
.θ .tan θ
.−1 .1
Notice
sin θ ≤ θ ≤ 2 tanθ
2≤ tan θ
Divideby sin θ:
1 ≤ θ
sin θ≤ 1
cos θ
Takereciprocals:
1 ≥ sin θ
θ≥ cos θ
As θ → 0, theleftandrightsidestendto 1. So, then, mustthemiddleexpression.
. . . . . .
ProofoftheCosineLimit
Proof.
1− cos θ
θ=
1− cos θ
θ· 1 + cos θ
1 + cos θ=
1− cos2 θ
θ(1 + cos θ)
=sin2 θ
θ(1 + cos θ)=
sin θ
θ· sin θ
1 + cos θ
So
limθ→0
1− cos θ
θ=
(limθ→0
sin θ
θ
)·(limθ→0
sin θ
1 + cos θ
)= 1 · 0 = 0.
. . . . . .
Trythese
Example
I limθ→0
tan θ
θ
I limθ→0
sin 2θ
θ
Answer
I 1I 2
. . . . . .
Trythese
Example
I limθ→0
tan θ
θ
I limθ→0
sin 2θ
θ
Answer
I 1I 2