52338-exercicios sobre limites

7/30/2019 52338-Exercicios Sobre Limites

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EXERCICIOS SOBRE LIMITES:

Calcule os seguintes limites.

1)

Soluo:

4

3

2

5lim

)2)(2(

)5)(2(lim

10

107lim

222

2

2=

+

=

+

=

+

x

x

xx

xx

x

xx

xxx

2)Soluo:

3)Soluo:

4)Soluo:

5)

Soluo:

6)Soluo:


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7)Calcule:

Soluo:

8) Calcule:Soluo:

9) Calcule:

Soluo:

0|x|

)x(sen.xlim

ento0,)senx(limcomo)Sandwich(ConfrontodoTeoremaPelo

sen(x)

|x|

x.sen(x)sen(x)-:temos1,

|x|

x1-quedesde

0x

x

=

=

10)Calcule5x

100lim

2x +

Soluo:

=

+

100

5x

100lim

2x

The numerator is always 100 and the denominator x2 + 5 approaches as x approaches , so that the resulting

fraction approaches 0.

11) Calcule20x

7lim

3x .

Soluo:

0-

7

20x

7lim

3x=

=

The numerator is always 7 and the denominator x3 7 approaches - as x approaches - , so that

the resulting fraction approaches 0.

12) Calcule: xlim

x

5

x

2

+ x 10Soluo:

Note that the expression x5 x2 leads to the indeterminate form - . Circumvent this by appropriatefactoring:

.As x approaches , each of the three expressions x2, (x3 1), and (x 10) approaches . Temos, ento:

.+

=

Thus, the limit does not exist. Note that an alternate solution follows by first factoring out , the highest

power ofx . Try it.

13) Calcule 5x3

7xlimx +

+

Soluo:


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=

+

+

5x3

7xlim

x

zeroatendemx

5

ex

7

,xquandoe

x

53

x

71

lim

x

5

x

x3

x

7

x

x

lim5x3

7x

lim

teremospor x,expressodardenominadoenumeradoroDividindo

xxx

+

+

=

+

+

=+

+

Desta forma,5x3

7xlim

x +

+

tende a

3

1

14)Calcule4x10x

7x3xlim

3

2

x ++

Soluo:

Note that the expression x2 3x leads to the indeterminate form - asx se approaches .Circumvent this by dividing each of the terms in the original problem by x3, the highest power ofx in

the problem .

When x the

4x10x

7x3xlim

3

2

x ++

approaches 0

= 0

15) Calcule 7x-xlim2

x+

Soluo:

7xx

7xxlim

7xx

)7xx)(7x-x(lim7x-xlim

2

22

x2

22

x

2

x

++

=

++

+++=+

=

7xx

7lim

2x ++

e quando

x , 7x-xlim2

x+

=7xx

7lim

2x ++

tende a zero

16) (Circumvent this indeterminate form by using the conjugate of the expression

Calculex

)xsin(limx

SOLUTION : First note that

because of the well-known properties of the sine function. Since we are computing the

limit asx goes to infinity, it is reasonable to assume thatx > 0 . Thus,

.


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Since

,

it follows from the Squeeze Principle that

17) Calcule3x

xcos2limx +

SOLUTION :First note that

because of the well-known properties of the cosine function. Now multiply by -1, reversing

the inequalities and getting

or

.

Next, add 2 to each component to get.

Since we are computing the limit asx goes to infinity, it is reasonable to assume that x + 3 > 0. Thus,

.

Since

,

it follows from the Squeeze Principle that

.

18) Calculex23

xcos2limx


because of the well-known properties of the cosine function, and therefore

.

Since we are computing the limit asx goes to infinity, it is reasonable to assume that3 - 2x < 0. Now divide each component by 3 - 2x, reversing the inequalities and getting

,or

.Since

,it follows from the Squeeze Principle that


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19) Calcule

x

2cosxlim 3

0x

SOLUTION : Note that

x

2cosxlim 3

0xDOES NOT EXIST since values of

2

xcos oscillate

between -1 and +1 asx approaches 0 from the left. However, this does NOT necessarily mean

that

x

2cosxlim 3

0xdoes not exist ! ? Indeed,x3 < 0 and

forx < 0. Multiply each component byx3, reversing the inequalities and getting

or

.

Since

,it follows from the Squeeze Principle that

.

20) Calcule

.


,

so that

and

.Since we are computing the limit asx goes to infinity, it is reasonable to assume that

x+100 > 0. Thus, dividing byx+100 and multiplying byx2, we get

and

.

Then

x

1001

x2lim

x

100xx

x2

lim100x

x2lim

x

2

x

2

x+

=+

=

+ quando x tende a

100x

)xsin2(xlim

22

x ++

igual a =

+

01

Similarly,

100xx3lim

2

x += .

Thus, it follows from the Squeeze Principle that


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100x

)xsin2(xlim

22

x ++

= (does not exist).

21) Calcule

.SOLUTION : First note that

,so that

,

,

and

.Then

= 5 .

Similarly,

510x

1x5lim

2

2

x=

++

Thus, it follows from the Squeeze Principle that

510x

)x3sin(x5lim

2

2

x=

+

.

22) Calcule


and

,so that

and

.Since we are computing the limit asx goes to negative infinity, it is reasonable to

assume thatx-3 < 0. Thus, dividing byx-3, we get

or


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.

Now divide byx2 + 1 and multiply byx2 , getting

.

Then

00001

0

x

3

x

1

x

31

x

2

lim

32

x=

+

=

+

Similarly,

0)3x)(1x(

x2lim

2

2

x=

+

It follows from the Squeeze Principle that

0)3x)(1x(

)xcosx(sinxlim

2

32

x=

++

LIMITES CONSTINUIDADE

1) Determine se a seguinte funo contnua emx=1 .

SOLUTION :: Functionfis defined atx = 1 since

i.)f(1) = 2 .

The limit

= 3 (1) - 5

= -2 ,i.e.,

ii.) 2)x(flim1x

=

.

But

iii.) )1(f)x(flim1x ,so condition iii.) is not satisfied and functionfis NOT continuous at x = 1.

2) Determine se a seguinte funo contnua emx = -2 .

SOLUTION : Functionfis defined atx=-2 sincef(-2) = (-2)2 + 2(-2) = 4-4 = 0 .

The left-hand limit

= (-2)2 + 2(-2)= 4 - 4

= 0 .


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The right-hand limit

= (-2)3 - 6(-2)= -8 + 12

= 4 .Since the left- and right-hand limits are not equal, ,

ii.) )x(flim2x does not exist,

and condition ii.) is not satisfied. Thus, functionfis NOT continuous atx = -2 .

3) Determine se a seguinte funo contnua emx = 0 .

SOLUTION : Functionfis defined atx = 0 since

i.)f(0) = 2 .The left-hand limit

= 2 .The right-hand limit

= 2 .

Thus,

)x(flim0x exists with

ii.) 2)x(flim0x

=

.

Since

iii.) )0(f2)x(flim0x

==

,

all three conditions are satisfied, andfis continuous atx=0 .

4) Determine se a funo1x

1x)x(h

3

2

+

+= contnua atx = -1 .

SOLUTION : Function h is not defined atx = -1 since it leads to division by zero.Thus, h (-1) does not exist, condition i.) is violated, and function h is NOT continuous

atx = -1 .

5) Check the following function for continuity atx = 3 andx = -3 .

SOLUTION : First, check for continuity atx=3 . Functionfis defined atx=3 since

.The limit

(Circumvent this indeterminate form by factoring the numerator and the denominator.)


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(Recall thatA2 -B2 = (A -B)(A +B) andA3 -B3 = (A -B)(A2 +AB +B2 ) . )

(Divide out a factor of (x - 3) . )

,

i.e.,

ii.) .

Since,

iii.) ,

all three conditions are satisfied, andfis continuous atx=3 . Now, check for continuity atx = -3 . Functionfis not defined atx = -3 because of division by zero. Thus,

i.)f(-3)does not exist, condition i.) is violated, andfis NOT continuous atx = -3 .

6) Para que valores dex a funo 4x3x

5x3x)x(f

2

2

+

++= contnua ?

SOLUTION 6 : Functionsy =x2 + 3x + 5 andy =x2 + 3x - 4 are continuous for allvalues ofx since both are polynomials. Thus, the quotient of these two functions,

4x3x

5x3x)x(f

2

2

+

++= , is continuous for all values ofx where the denominator,

y = x2 + 3x - 4 = (x - 1)(x + 4) , does NOT equal zero. Since (x - 1)(x + 4) = 0 forx = 1

andx = -4 , functionfis continuous for all values ofx EXCEPTx = 1 andx = -4 .

7) Para que valores dex a funo ( ) 31

20 )5xsin()x(g += contnua ?

SOLUTION: First describe functiongusing functional composition. Letf(x) =x1/3 ,

h(x) = sin(x), and k(x) =x20 + 5 . Function kis continuous for all values ofx since it is apolynomial, and functionsfand h are well-known to be continuous for all values ofx .

Thus, the functional compositions

and

are continuous for all values ofx . Since

,

functiongis continuous for all values ofx .

8) Para que valores dex a funo x2x)x(f 2 = contnua ?

SOLUTION : First describe functionfusing functional composition. Letg(x) =x2 - 2x

and h(x) = x . Functiongis continuous for all values ofx since it is a polynomial, and

function h is well-known to be continuous for 0x . Sinceg(x) =x2 - 2x =x(x-2) , itfollows easily that 0)x(g for 0x and 2x . Thus, the functional composition

is continuous for 0x and 2x and. Since

,


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functionfis continuous for 0x and 2x and.

9) Para que valores dex a funo

+

=2x

1xln)x(f contnua ?

SOLUTION : First describe functionfusing functional composition. Let

2x

1x)x(g

+

= and h(x) = ln(x). Sincegis the quotient of polynomialsy =x - 1 and

y = x + 2 , functiongis continuous for all values ofx EXCEPT wherex+2 = 0 , i.e.,EXCEPT forx = -2 . Function h is well-known to be continuous forx > 0 . Since

2x

1x)x(g

+

= , it follows easily thatg(x) > 0 forx < -2 andx > 1 . Thus, the functional

composition

is continuous forx < -2 andx > 1 . Since

,functionfis continuous forx < -2 andx > 1 .

10) Para que valores dex a funo 9x4

e)x(f

2

xsin

= contnua ?

SOLUTION 10 : First describe functionfusing functional composition. Let

g(x) = sin(x) and h(x) = ex, both of which are well-known to be continuous for all

values ofx . Thus, the numerator y = esin(x) = h(g(x)) is continuous (the functional

composition of continuous functions) for all values ofx . Now consider the denominator

9x4y 2 = . Letg(x) = 4 , h(x) =x2 - 9 , and x)x(k = . Functionsgand hAre continuous for all values ofx since both are polynomials, and it is well-known thatfunction kis continuous for 0x . Since h(x) =x2 - 9 = (x-3)(x+3) = 0 whenx = 3 or

x = -3 , it follows easily that -3xand3for x0)x(h for and, so that

))x(h(k9x4y 2 == is continuous (the functional composition of continuous

functions) for -3xand3x and. Thus, the denominator 9x4y 2 = is

continuous (the difference of continuous functions) for -3xand3x and. There

is one other important consideration. We must insure that the DENOMINATOR ISNEVER ZERO. If

then

.Squaring both sides, we get

16 =x2 - 9so that

x2 = 25when

x = 5 orx = -5 .Thus, the denominator is zero ifx = 5 orx = -5 . Summarizing, the quotient of these

continuous functions,9x4

e)x(f

2

xsin

= , is continuous for -3xand3x

and, but NOT forx = 5 andx = -5 .

Para que valores dex a seguinte funo contnua ?


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SOLUTION : Consider separately the three component functions which determinef.

Function1x

1xy

= is continuous forx > 1 since it is the quotient of continuous

functions and the denominator is never zero. Functiony = 5 -3x is continuous for

1x2 since it is a polynomial. Function4x

6y

= is continuous forx < -2 since it

is the quotient of continuous functions and the denominator is never zero. Now check forcontinuity offwhere the three components are joined together, i.e., check for continuity at

x = 1 andx = -2 . Forx = 1 functionfis defined sincei.)f(1) = 5 - 3(1) = 2 .


0

0

1x

1xlim)x(flim

1x1x=

=

++

(Circumvent this indeterminate form one of two ways. Either factor the numerator as the

difference of squares, or multiply by the conjugate of the denominator over itself.)

= 2 .

The left-hand limit

== 5 - 3(1)= 2 .

Thus,

ii.) 2)x(flim1x

=

.

Since

iii.) 2)x(flim1x

=

= f(1),

all three conditions are satisfied, and functionfis continuous atx = 1 . Now check forcontinuity atx = -2 . Functionfis defined atx = -2 since

i.)f(-2) = 5 - 3(-2) = 11 .The right-hand limit

=

= 5 - 3( -2)= 11 .

The left-hand limit

4x

6lim)x(flim

2x2x =

=

= -1 .

Since the left- and right-hand limits are different,

ii.) )x(flim2x

does NOT exist,

condition ii.) is violated, and functionfis NOT continuous atx=-2 . Summarizing,functionfis continuous for all values ofx EXCEPTx = -2 .


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12. Determine todos os valores da constanteA para que a seguinte funo seja contnua para todos os valores

dex .

SOLUTION : First, consider separately the two components which determine functionf.

Functiony =A2x -A is continuous for 3x for any value ofA since it is a polynomial.Functiony = 4 is continuous forx < 3 since it is a polynomial. Now determineA so that

functionfis continuous atx=3 . Functionfmust be defined atx = 3 , soi.)f(3)=A2 (3) -A = 3A2 -A .


)AxA(lim)x(flim 2

3x3x=

++

=A2 (3) -A

= 3A2 -A .The left-hand limit

4lim)x(flim3x3x

= = 4 .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) 4AA3)x(flim2

3x

==

,

so that= 3A2 -A - 4 = 0 .

Factoring, we get(3A - 4)(A + 1) = 0

for

34A = orA = -1 .

For either choice ofA ,

iii.) ,all three conditions are satisfied, andfis continuous atx = 3 . Therefore, functionfis

continuous for all values ofx if 34

A=

orA = -1 .

13. Determine todos os va;ores das constantes A e B para que a funo seja contnua para todos os valores de

x .

SOLUTION : First, consider separately the three components which determine functionf.

Functiony =Ax -B is continuous for 1x for any values ofA andB since it is apolynomial. Functiony = 2x2 + 3Ax +B is continuous for 1x1 for any values ofA

andB since it is a polynomial. Functiony = 4 is continuous forx > 1 since it is a polynomial.

Now determineA andB so that functionfis continuous atx=-1 andx=1 . First, considercontinuity atx = -1 . Functionfmust be defined atx = -1 , so

i.)f(-1)=A(-1) -B = -A -B .The left-hand limit

)BAx(lim)x(flim1x1x

=

=

=A (-1) -B= -A -B .


)BAx3x2(lim)x(flim 2

1x1x++=

++

=

= 2(-1)2 + 3A(-1) +B

= 2 - 3A +B .For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) BA32BA)x(flim1x

+== ,


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so that

2A - 2B = 2 ,or

(Equation 1)A -B = 1 .

Now consider continuity atx=1 . Functionfmust be defined atx=1 , soi.)f(1)= 2(1)2 + 3A(1) +B = 2 + 3A +B .

The left-hand limit)BAx3x2(lim)x(flim 2

1x1x++=

=

= 2(1)2 + 3A(1) +B

= 2 + 3A +B .The right-hand limit

4lim)x(flim1x1x++

= =

= 4 .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) 4BA32)x(flim1x

=++= ,

or

(Equation 2)3A +B = 2 .

Now solve Equations 1 and 2 simultaneously. Thus,

A -B = 1 and 3A +B = 2are equivalent to

A =B + 1 and 3A +B = 2 .Use the first equation to substitute into the second, getting

3 (B + 1 ) +B = 2 ,3B + 3 +B = 2 ,

and4B = -1 .

Thus,

and

.

For this choice ofA andB it can easily be shown that

iii.) )1(f4)x(flim1x

==

and

iii.) )1(f2

1)x(flim

1x==

,

so that all three conditions are satisfied at bothx=1 andx=-1 , and functionfis continuous at

bothx=1 andx=-1 . Therefore, functionfis continuous for all values ofx if

4-1Band

43A == and.

52338-exercicios sobre limites

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