. . . . . .

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