algebra of limits

Algebra of LimitsAssume that both of the following limits exist and c and is a real number:

Then:

0)(lim

),(lim/)(lim)(/)(lim..4

)(lim)(lim)()(lim..3

)(lim)(lim)()(lim..2

)(lim)(lim..1

xgthatprovided

xgxfxgxf

xgxfxgxf

xgxfxgxf

xfcxcf

ax

axaxax

axaxax

axaxax

axax

Calculating LimitsFinding the limit of a function f a point x = a.

Distinguishing the following cases:1. The case when f is continuous a x = a.2. The case 0/0.3. The case ∞/ ∞4. The case of an infinite limit5. The case of the function f defined by a formula

involving absolute values.6. The case c/∞, where c is a real number.7. Other cases: the case ∞- ∞8. The case, when it is possible to use the squeeze

theorem.

1. The case when f is continuous at x = a

If f is continues at x=a, then:

Notice:1. Polynomial functions and the cubic root function ( & all functions of its two families) are everywhere continuous.2. Rational, trigonometric and root functions are continuous at every point of their domains.3. If f and g are continuous a x=a, then so are cf, f+g, f-g, fg and f/g (provided that he limit of f at x=a is not zero)

Examples for the case when f is continuous at x = a

040

224)2()2()(lim

2.},2{

.24)(

24lim

)1(

2

2

2

2

2

fxf

xatcontisitsoandRon

contisxxxffunctionrationalThe

xx

Example

x

x

Examples for the case when f is continuous at x = a

22)2(248)321(

3191)3)1(2)1(()1()(lim

,.139)32()(

,1.),,3(3)(

1.9)32()(

1.,9)(

1.,32)(

39)32(lim

)2(

9

95

1

95

95

9

5

95

1

fxf

soandxatcontisxxxxxffunctionThe

ThusxatcontisitthusoncontisxxhfunctionrootThe

xatcontisxxxxgfunctionproducttheTherefore

xatcontisitthuseverywherecontisxxsfunctionThe

xatcontisitthuseverywherecontisxxxpfunctionpolynomialThe

xxxx

Example

x

x

Examples for the case when f is continuous at x= a

155515)5(53lim

5.53)(:,5.

).25..1

:(

.5)(3)(

53lim:)3(

5

5

fxxThus

xatcontisxxxfthusxatcontaretheysoand

hGraphxatcontishthatShow

Questions

everywherecontarexxhandxxg

xxExample

x

x

2. The case 0/0

Suppose we want to find:

For the case when:

Then this is called the case 0/0. Caution: The limit is not equal 0/0. This is just a name that classifies the type of limits having such property.

)()(limxhxg

ax

.0)(lim&)(lim arexhxgaxax

Examples for the case 0/0

83

3212

)8(4444

)4)(2(42lim

)4)(2)(2()42)(2(lim

168lim

16lim08lim

:,0/0168lim

:)1(

2

2

2

2

2

2

4

3

2

4

2

3

2

4

3

2

xxxx

xxxxxx

xx

xx

becausecasetheisThisxx

factoringbySolvingExample

x

x

x

xx

x

Examples for the case 0/0

301

)55(31

525)3(1lim

525)3(lim

525)3(25)25(lim

525525.

)3(525lim

)3(525lim

)3(lim0525lim

:,0/0)3(525lim

)2(

0

00

00

00

0

xx

xxxx

xxxx

xx

xxx

xxx

xxx

becausecasetheisThisxx

x

methodconjugatethebygMultiplyinExample

x

xx

xx

xx

x

3. The case ∞/ ∞


For the case when the limits of both functions f and g are infinite

Then this is called the case ∞/ ∞. Caution: The limit is not equal ∞/∞. This is just a name that classifies the type of limits having such property.

)()(lim

)()(lim

xhxgOr

xhxg

xax

Examples for the case ∞/∞

1. See the examples involving rational functions in the file on the limit a infinity.

2. Examples involving roots: See the following slides

Limits at infinity

A function y=f(x) may approach a real number b as x increases or decreases with no bound.When this happens, we say that f has a limit at infinity, and that the line y=b is a horizontal asymptote for f.

1. Limit at infinity: The Case of Rational Functions

A rational function r(x) = p(x)/q(x) has a limit at infinity if the degree of p(x) is equal or less than the degree of q(x).

A rational function r(x) = p(x)/q(x) does not have a limit at infinity (but has rather infinite right and left limits) if the degree of p(x) is greater than the degree of q(x).

Example (1)Let

Find

Solution:Since the degree of the polynomial in the numerator, which is 9, is equal to the degree of the polynomial in the denominator, then

146325)( 79

29

xxxxxf

)(lim xfx

To show that, we follow the following steps:

65

)(lim 9

9

inatordenomtheinxofcofficientThe

numeratortheinxofcofficientThexfx

65

0)0(46)0(3)0(25

1lim1lim46lim

1lim31lim25lim

1146lim

3125lim

1146

3125lim

146325lim)(lim

92

97

92

97

92

97

79

29

xx

xx

xx

xx

xx

xxxxxxxf

xxx

xxx

x

x

xxx

Example (2)Let

Find

Solution:Since the degree of the polynomial in the numerator, which is 9, is less than the degree of the polynomial in the denominator, which is 12, then

146325)( 712

29

xxxxxf

)(lim xfx

To show that, we follow the following steps:

0)(lim

xfx

060

0)0(46)0(3)0(2)0(5

1lim1lim46lim

1lim31lim21lim5

1146lim

3125lim

1146

3125

lim146325lim)(lim

125

12103

125

12103

125

12103

712

29

xx

xxx

xx

xxx

xx

xxxxxxxxf

xxx

xxx

x

x

xxx

Example (3)Let

Find

Solution:Since the degree of the polynomial in the numerator, which is 12, is greater than the degree of the polynomial in the denominator, which is 9, then

146325)( 79

212

xxxxxf

)(lim xfx

They are infinite limits. To show that, we follow the following steps:

3

9

12

79

212

lim65lim

146325lim)(lim

xassamethearewhichxxassametheare

xxxxxf

x

x

xx

.)(lim existnotdoxfx

Example (4)Let

Find


146325)( 79

212

xxxxxf

)(lim xfx


)(lim65lim

146325lim)(lim

3

9

12

79

212


xxxxxf

x

x

xx


Example (5)Let

Find


146325)( 78

212

xxxxxf

)(lim xfx



4

8

12

78

212

lim65lim

146325lim)(lim


xxxxxf

x

x

xx

Example (6)Let

Find


146325)( 78

212

xxxxxf

)(lim xfx



)(lim65lim

146325lim)(lim

4

8

12

78

212


xxxxxf

x

x

xx

Limits @ Infinity

2. Problems Involving Roots

Introduction

We know that:

√x2 = |x|, which is equal x is x non-negative and equal to – x if x is negative

For if x = 2, then √(2)2 = √4 = |2|=2

& if x = - 2, then √(-2)2 = √4 = |-2|=-(-2) = 2

Example

0;16

0;30;16

9

92

92

2

2

29

29

2

2

2

16

16

16

916)(

:916)(

:

xx

xxx

x

x

x

x

x

x

x

x

xxf

asrewwrittenbecanxxf

Let

Example (1)

fofasymptoteshorizontaltheareyandylinestheThusWhy

x

x

xx

xx

xxxf

atandatfofitsmlithefrstfindWeasymptoteshorizontalFinding

Solution

fofasymptotesallFindxxxf

Let

xxxx

22?2

)22(

916lim

)22(

916lim

22916lim)(lim

:.1:

22916)(

:

222

2

)1(2916lim)(lim

,.015916,1

:1,)1(2916lim)(lim

,.015916,1

1

:.122916)(

2

11

2

2

11

2

2

xxxf

Thusnegativekeepingwhileapproachesxandapproachesxleftthefromapproachesxas

asymptoteverticalaisxhencexxxf

Thuspositivekeepingwhileapproachesxandapproachesxrightthefromapproachesxas

numeratorthenotandnatordomitheofzeroaisxmitlifinitenianhasfwherefindfirstWe

asymptotesverticalFindingxxxf

xx

xx

)23,0(int

23

2090)0(

potheataxisythewithectionInters

f

Example (2)

fofasymptoteshorizontaltheareyandylinestheThus

x

x

xx

xx

xxxf

exampleprevioustheindidweasatandatfofitsmlithefindfrstWeasymptoteshorizontalFinding

Solution

itgraphfofasymptotesallFindxxxf

Let

xxxx

222

)22(

916lim

)22(

916lim

22916lim)(lim

)(:.1

:

!&22916)(

:

222

2

)1(2916lim)(lim

,.015916,1

:1,)1(2916lim)(lim

,.015916,1

1

:.122916)(

2

11

2

2

11

2

2

xxxf

Thusnegativekeepingwhileapproachesxandapproachesxleftthefromxapproachesxas

asymptoteverticalaisxhencexxxf

Thuspositivekeepingwhileapproachesxandapproachesxrightthefromapproachesxas

numeratorthenotandnatordomitheofzeroaisxmitlifinitenianhasfwherefindfirstWe

asymptotesverticalFindingxxxf

xx

xx

Homework:Problems:Example (4) & (5) –Section 3.4. Page: 228Exercises 3.4 Page: 235 :Problems: 9, 10, 11, 15, 23, 25,17 & 40

4. The case of infinite limit

See the examples in the file on the infinite limits and also the examples of infinite limits in the file on limits at infinity.

Infinite Limits

A function f may increases or decreases with no bound near certain values c for the independent variable x. When this happens, we say that f has an infinite limit, and that f has a vertical asymptote at x = c The line x=c is called a vertical asymptote for f.

Infinite Limits- The Case of Rational Functions

A rational function has an infinite limit if the limit of the denominator is zero and the limit of the numerator is not zero. The sign of the infinite limit is determined by the sign of both the numerator and the denominator at values close to the considered point x=c approached by the variable x.

Example (1)Let

Find

Solution:First x=0 is a zero of the denominator which is not a zero of the numerator.

xxf 1)(

)(lim.0

xfax

)(lim.0

xfbx

a. As x approaches 0 from the right, the numerator is always positive ( it is equal to 1) and the denominator approaches 0 while keeping positive; hence, the function increases with no bound. Thus:

)(lim0

xfx

The function has a vertical asymptote at x = 0, which is the line x = 0 (see the graph in the file on basic algebraic functions).

b. As x approaches 0 from the left, the numerator is always positive ( it is equal to 1) and the denominator approaches 0 while keeping negative; hence, the function decreases with no bound.

)(lim0

xfx

Example (2)Let

Find

Solution:First x=1 is a zero of the denominator which is not a zero of the numerator.

)(lim.1

xfax

)(lim.1

xfbx

15)(

xxxf

)(lim1

xfx

The function has a vertical asymptote at x = 1, which is the line x = 1

)(lim1

xfx

a. As x approaches 1 from the right, the numerator approaches 6 (thus keeping positive), and the denominator approaches 0 while keeping positive; hence, the function increases with no bound. Thus,

b. As x approaches 1 from the left, the numerator approaches 6 (thus keeping positive), and the denominator approaches 0 while keeping negative; hence, the function decreases with no bound. The function has a vertical asymptote at x=1, which is the line x = 1. Thus:

Example (3)Let

Find Solution:First, rewrite: x=0 is a zero of the denominator which is not a zero of the numerator.

)(lim.0

xfax

)(lim.0

xfbx

54

11)(xx

xf

554

111)(xx

xxxf

)(lim0

xfx

The function has a vertical asymptote at x = 0, which is the line x = 0

)(lim0

xfx

a. As x approaches 0 from the right, the numerator approaches -1 (thus keeping negative), and the denominator approaches 0 while keeping positive; hence, the function decreases with no bound. Thus:

b. As x approaches 0 from the left, the numerator approaches -1 (thus keeping negative), and the denominator approaches 0 while keeping negative; hence, the function increases with no bound. Thus:

Same Type Problems from the Homework

Exercises 1.5 Pages 59-61Problems: 29, 31, 33, 37

5. The case of discontinuous function f defined by a formula involving absolute values

fGraphQuestion

existnotdoesxx

xx

xx

xx

xffunctionThe

Solutionxx

existsifFindExample

x

xx

xx

xx

xxx

x

:5153

lim

5153

lim335153

lim

5153

)(

:5153

lim

:,:)1(

5

55

5;35;3

5;5153

5;5

)153(

5

5. The case of discontinuous function f defined by a formula involving absolute values

existnotdoesxx

xfx

xf

xxxffunctionThe

Solution

xx

existsifFindExample

x

xxxx

xx

x

xxx

xxx

x

21

21lim

00lim)(lim,1lim)(lim

21

21)(

:

21

21lim

:,:)2(

0

0000

0;1

0;0

0;21

21

0;21

21

0


Exercises 1.6 Pages 69-71Problems: 43, 45, 42, 44

6. The case constant/∞Suppose we want to find:

For the case when:

In this case, no mater what the formulas of g and h are, we will always have:

Then this is called the case c/∞. Caution: The limit is not equal c/ ∞. This is just a name that classifies the type of limits having such property. This limit is always equal zero

)()(limxhxg

ax

)(lim&)(lim xhRcxgaxax

0)()(lim

xhxg

ax

Example on the case constant/∞

011lim

)()(lim

,

?)()(lim1)(

11lim)(lim1)(::11lim

,:)1(

xxxhxg

Thus

Whyxhxxxh

xgxghaveWe

Solutionxx

FindExample

xx

x

xx

x

Same Type Problems

36

24

5

23lim.2

1

25lim.1

,:)1(

xx

xxx

FindExample

x

x

7. Other cases: the case ∞- ∞, 0∙∞


Then this is called the case ∞ - ∞. Caution: The limit is not equal ∞ - ∞. This is just a name that classifies the type of limits having such property.

arebothorarehandgfunctionsbothofmitslithewhen

xhxgOrxhxgxax

)()(lim)()(lim

Example for the case ∞- ∞

casethehaveweThus

xxhxxh

xxxxg

xx

xxxxg

Discussion

xx

existsifFindExample

xx

xxx

xx

x

x

xx

x

x

,

lim)(lim)(

71lim7lim)(lim

71717)(

:

7lim

:,:)1(

22

0;71

0;7

0;71

2222

2

2

2

?07

7lim

7

)7(lim

7

77lim7lim

2

2

22

2

222

Whyxx

xx

xxxx

xxxxxx

x

x

xx

121

12

1lim)2)(1(

1lim

0/0,23

1lim231)2(lim

)2)(1(1

11lim

231

11lim

,23

1lim&1

1lim

231lim&

11lim

::

231

11lim

:,:)2(

11

2121

121

211

211

21

xxxx

casethehaveweNowxx

xxx

x

xxxxxx

casethehaveweThusxxx

xxx

haveWeSolution

xxx

existsifFindExample

xx

xx

xx

xx

xx

x

)21:(

12

11lim.1

)21:(1

11lim.1

:,:

21

0

Answerxx

Answerxxx

existifFindPeoblems

x

x

7. Using the Squeeze Theorem

The Squeeze (Sandwich or Pinching)) Theorem

Suppose that we want to find the limit of a function f at a given point x=a and that the values of f on some interval containing this point (with the possible exception of that point) lie between the values of a couple of functions g and h whose limits at x=a are equal. The squeeze theorem that in this case the limit of f at x=a will equal the limit of g and h at this point.

The Squeeze Theorem

lxfThen

xhlxg

dcawhereadcxxhxfxg

Let

ax

axax

)(:

)()(&

),(),(;)()()(

:

lim

limlim

Example (1)

1)(,,

)4,0(;)(12,)4,0(1

&

1)1(&112)12(

::

)(

)4,0(;)(12

:

lim

limlim

lim

1

2

22

11

1

2

xftheoremsqueezethebyThus

xxxfx

xx

haveWeSoluion

xfFind

xxxfx

Let

x

xx

x

Example (2)

7)(,,

),0[;)(12,),0(2

&

771616)74(

,7916)94(::

)(

),0[;74)(94

:

lim

limlim

lim

4

2

2

4

4

4

2

xftheoremsqueezethebyThus

xxxfx

xx

xhaveWe

Soluion

xfFind

xxxxfx

Let

x

x

x

x

Example (3)

0)1sin(

,

)(0)(

'

0)5,5(;1sin

)...(

?0)5,5(;11sin1

::

:

)1sin(

2

0

2

0

2

0

222

22

2

0

lim

limlim

lim

xx

theoremsqueeztheBy

xx

haveWe

xxx

xx

negativenonisxthatNotexbygMultiplyin

Whyxx

haveWe

Soluion

xx

Find

x

xx

x

Example (4)

0)2cos(

,

)(0)(

'

0)5,5(;2cos

)...(

?0)5,5(;12cos1

::

:

)2cos(

4

0

4

0

4

0

444

44

4

0

lim

limlim

lim

xx

theoremsqueezetheBy

xx

haveWe

xxx

xx

negativenonisxthatNotexbygMultiplyin

Whyxx

haveWe

Soluion

xx

Find

x

xx

x


Exercises 6.1 Pages 70Problems: 37, 38, 39

algebra of limits

Documents