resolução limites

8/2/2019 Resoluo Limites

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Resoluo Limites

a)

b)

c)

d)

e)

f)


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g)

h)

i)

j)

LIMITES DE FUNES QUANDO X SE APROXIMA DE UMA CONSTANTE

.

O seguintes problemas requerem o uso do clculo de limites de funes ao x se aproximar deuma constante. A maioria dos problemas de nvel mdio. Alguns so engenhosos. Todas as soluesso dadas sem o uso da regra de L'Hopital


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PROBLEMA 1 :Calcule

.

o PROBLEMA 2 :Calcule

.


.


.


.


.



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.


.


.


o .o PROBLEMA 11 :Calcule

.


.



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.

Este PROBLEMA requer uma substituio especial, identidades trigonomtricas, elimites trigonomtricos.

O seguinte problema requer a noo de limites laterais

o PROBLEMA 14 :Considere a funo

i.) Desenhe o grfico de f.

ii.) Determine os sguintes limites.

a.)

b.)

c.)

d.) e.)

f.)

g.)

h.)

i.)

j.)

k.)

l.)o PROBLEMA 15 :Consider the function


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Determine os valores das constantes ae btal que existe e seja igual a f(2).

SOLUTIONS TO LIMITS OF FUNCTIONS AS X APPROACHES ACONSTANT

SOLUTION 1 :

.

ClickHERE to return to the list of problems.

SOLUTION 2 :

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

(Divide out the factorsx - 2 , the factors which are causing the indeterminate form . Now
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%201http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%201http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%201http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%201


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the limit can be computed. )


SOLUTION 3 :

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

(Divide out the factorsx - 3 , the factors which are causing the indeterminate form . Nowthe limit can be computed. )


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.


SOLUTION 4 :

(Algebraically simplify the fractions in the numerator using a common denominator.)

(Division by is the same as multiplication by .)

(Factor the denominator . Recall that .)

(Divide out the factorsx + 2 , the factors which are causing the indeterminate form . Nowthe limit can be computed. )


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.


SOLUTION 5 :

(Eliminate the square root term by multiplying by the conjugate of the numerator over itself.Recall that

. )



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.


SOLUTION 6 :

(It may appear that multiplying by the conjugate of the numerator over itself is a reasonablenext step.

It's a good idea, but doesn't work. Instead, writex - 27 as the difference of cubes and recall that

.)

(Divide out the factors , the factors which are causing the indeterminate form .Now the limit can be computed. )


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= 27 .


SOLUTION 7 :

(Multiplying by conjugates won't work for this challenging problem. Instead, recall that

and , and

note that and . This should help explain the nextfew mysterious steps.)



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.


SOLUTION 8 :

(If you wrote that , you are incorrect. Instead, multiply and divide by 5.)

(Use the well-known fact that .)

.

http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%207http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%207http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%207http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%208http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%208http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%208http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%208http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%207


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SOLUTION 9 :

(Recall the trigonometry identity .)

(The numerator is the difference of squares. Factor it.)

(Divide out the factors , the factors which are causing the indeterminate form .Now the limit can be computed. )

.



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SOLUTION 10 :

(Factorx from the numerator and denominator, then divide these factors out.)

(The numerator approaches -7 and the denominator is a positve quantity approaching 0 .)

(This is NOT an indeterminate form. The answer follows.)

.

(Thus, the limit does not exist.)


SOLUTION 11 :

(The numerator approaches -3 and the denominator is a negative quantity which approaches 0 asx

approaches 0 .)


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(This is NOT an indeterminate form. The answer follows.)

.



SOLUTION 12 :

(Recall that . )


.

(The numerator approaches 1 and the denominator approaches 0 as x approaches 1 . However,

the quantity

in the denominator is sometimes negative and sometimes positive. Thus, the correct answer isNEITHER

NOR . The correct answer follows.)


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The limit does not exist.


SOLUTION 13 :

(Make the replacement so that . Note that

asx approaches , h approaches 0 . )

(Recall the well-known, but seldom-used, trigonometry

identity .)

(Recall the well-known trigonometry identity . )


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(Recall that . )

= 2 .


The next problem requires an understanding of one-sided limits.

SOLUTION 14 : Consider the function

i.) The graph offis given below.


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ii.) Determine the following limits.

a.) .

b.) .

c.) We have that does not exist since does not equal .

d.) .

e.) .

f.) We have that since .

g.) We have that (The numerator is

always -1 and the denominator is always a positive number approaching 0.) , so the limit does

not exist.

h.) .

i.) We have that does not exist since does not equal .

j.) .


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k.) .

l.) .


SOLUTION 15 : Consider the function

Determine the values of constants a and b so that exists. Begin by computing one-

sided limits atx=2 and setting each equal to 3. Thus,

and

.

Now solve the system of equations

a+2b = 3 and b-4a = 3 .

Thus,

a = 3-2b so that b-4(3-2b) = 3

iffb-12+ 8b = 3

iff 9b = 15

iff .

Then


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LIMITES DE FUNES QUANDO X SE APROXIMA DO INFINITO

Os seguintes problemas requerem o clculo algbrico de limites de funes ao

x se aproximar de + ou - infinito. Alguns so engenhosos. Todas as solues soobtidas sem a regra de L'Hopital. Tente obter suas solues sem olhar as solues

dadas levando cuidadosamente em considerao as

formas e durante seus clculos. Inicialmente alguns estudantes

incorretamante concluem que igual a 1 , ou que o limite no existe, ou

ou . Muitos concluem que igual a 0. De fato, as

formas e so exemplos de formas indeterminadas . Isto significa que

voc ainda no determinou uma resposta. Usualmente estas formas indeterminadas

podem ser contornadas por manipulaes algbricas. As solues esto todas em

ingls.

o PROBLEMA 1 : Calcule

.

o

PROBLEMA 2 :

Calcule

.


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.

o PROBLEMA 4 :

.

o PROBLEMA 5 :

.

o PROBLEMA 6 :

.


.


.

SOLUTIONS TO LIMITS OF FUNCTIONS AS X APPROACHES PLUS ORMINUS INFINITY


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SOLUTION 1 :

= = 0 .

(The numerator is always 100 and the denominator approaches as x approaches ,so that the resulting fraction approaches 0.)


SOLUTION 2 :

= = 0 .

(The numerator is always 7 and the denominator approaches as xapproaches , so that the resulting fraction approaches 0.)


SOLUTION 3 :

=

(This is NOT equal to 0. It is an indeterminate form. It can be circumvented by factoring.)

=

(As x approaches , each of the two expressions and 3x - 100 approaches .)

=
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(This is NOT an indeterminate form. It has meaning.)

= .

(Thus, the limit does not exist. Note that an alternate solution follows by first factoringout , the highest power ofx . Try it.)


SOLUTION 4 :

=

(As x approaches , each of the two expressions and approaches . )

=


= .

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

out , the highest power ofx . Try it.)


SOLUTION 5 :

(Note that the expression leads to the indeterminate form . Circumventthis by appropriate factoring.)

= .


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(As x approaches , each of the three expressions , , andx - 10 approaches .)

=

=

= .

(Thus, the limit does not exist. Note that an alternate solution follows by first factoringout , the highest power ofx . Try it. )


SOLUTION 6 :

=

(This is an indeterminate form. Circumvent it by dividing each term byx .)

=

=

=
http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%205http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%205http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%205http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%205


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(Asx approaches , each of the two expressions and approaches 0.)

=

= .


SOLUTION 7 :

(Note that the expression leads to the indeterminate form as xapproaches . Circumvent this by dividing each of the terms in the original problem by .)

=

=

=


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(Each of the three expressions , , and approaches 0 asx approaches .)

=

= .


SOLUTION 8 :

(Note that the expression leads to the indeterminate

form asx approaches . Circumvent this by dividing each of the terms in theoriginal problem by , the highest power ofx in the problem . This is not the only step thatwill work here. Dividing by , the highest power ofx in the numerator, also leads to the

correct answer. You might want to try it both ways to convince yourself of this.)

=

=


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=

(Each of the five expressions , , , , and approaches 0 asx approaches .)

=

= 0 .


SOLUTION 9 :

(Note that the expression leads to the indeterminate

form asx approaches . Circumvent this by dividing each of the terms in theoriginal problem by , the highest power ofx in the problem. . This is not the only step thatwill work here. Dividing byx , the highest power ofx in the denominator, actually leads moreeasily to the correct answer. You might want to try it both ways to convince yourself of this.)

=

=


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=

(Each of the three expressions , , and approaches 0 asx approaches .)

=

=

(This is NOT an indeterminate form. It has meaning. However, to determine it's exactmeaning requires a bit more analysis of the origin of the 0 in the denominator. Note

that = . It follows that ifx is a negative number then both of the

expressions and are negative so that is positive. Thus, for the

expression the numerator approaches 7 and the denominator is a positivequantity approaching 0 asx approaches . The resulting limit is .)

= .



SOLUTION 10 :


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=

(You will learn later that the previous step is valid because of the continuity of the square rootfunction.)

=

(Inside the square root sign lies an indeterminate form. Circumvent it by dividing each termby , the highest power ofxinside the square root sign.)

=

=

=

(Each of the two expressions and approaches 0 asx approaches .)

=

=


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= .


SOLUTION 11 :

= `` ''

(Circumvent this indeterminate form by using the conjugate of the

expression in an appropriate fashion.)

=

(Recall that .)

=

=

=

=

= 0 .



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SOLUTION 12 :

=


= .


CONTINUIDADE DE FUNES DE UMA VARVEL

Os seguintes problemas envolvem a continuidade de funes de uma varivel.Umafunoy =f(x) contnua em um pontox=ase as seguintes trs condies soverseicadas:

i.)f(a) est definida,

ii.) exte (i.e., finito) ,

e

iii.) .

Uma funof dita ser contnua no intervaloIsef contnua em cadapontox emI. Aqui est uma lista de alguns fatos bem conhecidos relacionados acontinuidade:

1. A SOMA de funes contnuas contnua.

2. A DIFERENA de funes contnuas contnua.

3. O PRODUTO de funes contnuas contnua.

4. O QUOCIENTE de funes contnuas contnua em todos os


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pontosx aonde o DENOMINADOR NO ZERO.

5. A COMPOSIO de funes contnuas contnua em todos ospontosx onde a composio est propriamente definida.

6. Qualquer polinmio contnuo para todos os valores dex.

7. A funo ex e as funes trigonomtricas e so contnua paratodos os valores dex.

o PROBLEMA 1 : Determine se a seguinte funo contnua emx=1 .

o PROBLEMA 2 : Determine se a seguinte funo contnua emx=-2 .

o PROBLEMA 3 : Determine se a seguinte funo contnua emx=0 .

o PROBLEMA 4 : Determine se a funo contnua atx=-1 .


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o PROBLEMA 5 : Check the following function for continuity atx=3 andx=-3 .

o PROBLEMA 6 : Para que valores dexa funo contnua ?

o PROBLEMA 7 : Para que valores dexa funo

contnua ?

o

PROBLEMA 8 :

Para que valores dex

a funo contnua ?

o PROBLEMA 9 : Para que valores dexa funo contnua ?

o PROBLEMA 10 : Para que valores dexa funo contnua?


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o PROBLEMA 11 : Para que valores dex aseguinte funo contnua ?

o PROBLEMA 12 : Determine todos os valores da constanteA para que a seguinte

funo seja contnua para todos os valores dex.

o PROBLEMA 13 : Determine todos os va;ores das constantes A e B para que a funo

seja contnua para todos os valores dex.

o PROBLEMA 14 : Mostre que a seguinte funo contnua para todos os valores

dex.


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o PROBLEMA 15 : Seja

Mostre quef contnua para todos os valores dex . Mostre quefdiferencivel para todos os valores dex, mas que a derivada ,f' , NO contnua emx=0 .


.


.


.


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.


.


.


.


.


.


.


.


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ii.) .

But

iii.) ,

so condition iii.) is not satisfied and functionfis NOT continuous atx=1 .


SOLUTION 2 : Functionfis defined atx=-2 since

i.)f(-2) = (-2)2 + 2(-2) = 4-4 = 0 .

The left-hand limit

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

= 4 - 4

= 0 .

The right-hand limit

= (-2)3 - 6(-2)

= -8 + 12

= 4 .

Since the left- and right-hand limits are not equal, ,
http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%201http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%201http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%201http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%201


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ii.) does not exist,

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


SOLUTION 3 : Functionfis defined atx=0 since

i.)f(0) = 2 .

The left-hand limit

= 2 .


= 2 .

Thus, exists with

ii.) .

Since


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iii.) ,

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


SOLUTION 4 : Function h is not defined atx=-1 since it leads to division by zero.Thus,

i.) h(-1)

does not exist, condition i.) is violated, and function h is NOT continuous atx = -1 .


SOLUTION 5 : First, check for continuity atx=3 . Functionfis defined atx=3 since

i.) .

The limit

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

(Recall thatA2 -B2 = (A-B)(A+B) andA3 -B3 = (A-B)(A2+AB+B2 ) . )
http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%203http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%203http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%203http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%204http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%204http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%204http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%204http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/continua/Continuity.html#PROBLEM%203


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

=

,

i.e.,

ii.) .

Since,

iii.) ,

all three conditions are satisfied, andfis continuous atx=3 . Now, check forcontinuity 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 .


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


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functions, , is continuous for all values ofx where thedenominator,y=x2 + 3x - 4 = (x-1)(x+4) , does NOT equal zero. Since (x-1)(x+4) = 0forx=1 andx=-4 , functionfis continuous for all values ofx EXCEPTx=1 andx=-4 .


SOLUTION 7 : First describe function g using functional composition. Letf(x)

=x1/3 , , and k(x) =x20 + 5 . Function kis continuous for all valuesofx since it is a polynomial, and functionsfand h are well-known to be continuousfor all values ofx . Thus, the functional compositions

and

are continuous for all values ofx . Since

,

function g is continuous for all values ofx .


SOLUTION 8 : First describe functionfusing functional composition. Let g(x) =x2 -

2x and . Function g is continuous for all values ofx since it is a

polynomial, and function h is well-known to be continuous for . Since g(x)

=x2 - 2x =x(x-2) , it follows easily that for and . Thus, thefunctional composition


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is continuous for and . Since

,

functionfis continuous for and .


SOLUTION 9 : First describe functionfusing functional composition.

Let and . Since g is the quotient of polynomialsy =x-1andy =x+2 , function g is 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 , it follows easily that g(x) > 0 forx < -2 andx > 1 . Thus, thefunctional composition

is continuous forx < -2 andx > 1 . Since

,

functionfis continuous forx < -2 andx > 1 .


SOLUTION 10 : First describe functionfusing functional composition.


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

values ofx . Thus, the numerator is continuous (the functionalcomposition of continuous functions) for all values ofx . Now consider the

denominator . Let g(x) = 4 ,h(x) =x2 - 9 , and .Functions g and h are continuous for all values ofx since both are polynomials, and it

is well-known that function kis continuous for . Since h(x) =x2 - 9 = (x-

3)(x+3) = 0 whenx=3 orx=-3 , it follows easily that for and ,

so that is continuous (the functional composition of

continuous functions) for and . Thus, the

denominator is continuous (the difference of continuous functions)

for and . There is one other important consideration. We must insurethat the DENOMINATOR IS NEVER ZERO. If

then

.

Squaring both sides, we get

16 =x2 - 9

so that

x2 = 25

when

x = 5 orx = -5 .

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


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continuous functions, , is continuous for and , butNOT forx = 5 andx = -5 .


SOLUTION 11 : Consider separately the three component functions which

determinef. Function is continuous forx > 1 since it is the quotient ofcontinuous functions and the denominator is never zero. Functiony = 5 -3x is

continuous for since it is a polynomial. Function is continuousforx < -2 since it is the quotient of continuous functions and the denominator isnever zero. Now check for continuity offwhere the three components are joinedtogether, i.e., check for continuity atx=1 andx=-2 . Forx = 1 functionfis definedsince

i.)f(1) = 5 - 3(1) = 2 .


=

(Circumvent this indeterminate form one of two ways. Either factor the numerator asthe difference of squares, or multiply by the conjugate of the denominator overitself.)


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= 2 .

The left-hand limit

=

= 5 - 3(1)

= 2 .

Thus,

ii.) .

Since

iii.) ,

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 .


=

= 5 - 3( -2)

= 11 .

The left-hand limit

=


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= -1 .

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

ii.) does NOT exist,

condition ii.) is violated, and functionfis NOT continuous atx=-2 . Summarizing,

functionfis continuous for all values ofxEXCEPTx=-2 .


SOLUTION 12 : First, consider separately the two components which determine

functionf. Functiony =A2x -A is continuous for for any value ofA since it isa polynomial. Functiony = 4 is continuous forx < 3 since it is a polynomial. NowdetermineA so that functionfis continuous atx=3 . Functionfmust be definedatx=3 , so

i.)f(3)=A2 (3) -A = 3A2 -A .


=

=A2 (3) -A

= 3A2 -A .

The left-hand limit

=


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= 4 .

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

ii.) ,

so that

3A2 -A - 4 = 0 .

Factoring, we get

(3A - 4)(A + 1) = 0

for

orA = -1 .

For either choice ofA ,

iii.) ,

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

continuous for all values ofx if orA = -1 .


SOLUTION 13 : First, consider separately the three components which determine

functionf. Functiony =Ax -B is continuous for for any valuesofA andB since it is a polynomial. Functiony = 2x2 + 3Ax +B is continuous for

for any values ofA andB since it is a polynomial. Functiony = 4 iscontinuous forx > 1 since it is a polynomial. Now determineA andB so thatfunctionfis continuous atx=-1 andx=1 . First, consider continuity atx=-1 .Functionfmust be defined atx=-1 , so


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

The left-hand limit

=

=A (-1) -B

= -A -B .


=

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

= 2 - 3A +B .


ii.) ,

so that

2A - 2B = 2 ,

or

(Equation 1)

A -B = 1 .

Now consider continuity atx=1 . Functionfmust be defined atx=1 , so

i.)f(1)= 2(1)2 + 3A(1) +B = 2 + 3A +B .

The left-hand limit

=


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

= 2 + 3A +B .


=

= 4 .


ii.) ,

or

(Equation 2)

3A +B = 2 .

Now solve Equations 1 and 2 simultaneously. Thus,

A -B = 1 and 3A +B = 2

are 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 ,

and

4B = -1 .

Thus,


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and

.

For this choice ofA andB it can easily be shown that

iii.)

and

iii.) ,

so that all three conditions are satisfied at bothx=1 andx=-1 , and functionfiscontinuous at bothx=1 andx=-1 . Therefore, functionfis continuous for all values

ofx if and .


SOLUTION 14 : First describefusing functional composition. Let g(x) = -1/x2 and h(x) = ex . Function h is well-known to be continuous for all values ofx .Function g is the quotient of functions continuous for all values ofx , and is thereforecontinuous for all values ofx exceptx=0 , thatx which makes the denominator zero.Thus, for all values ofx exceptx=0 ,

f(x) = h ( g(x) ) = eg(x)

= e-1/x2

is a continuous function (the functional composition of continuous functions). Nowcheck for continuity offatx=0 . Functionfis defined atx=0 since

i.)f(0) = 0 .


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The limit

(The numerator approaches -1 and the denominator is a positive number approachingzero.)

,

so that

= 0 ,

i.e.,

ii.) .

Since

iii.) ,

all three conditions are satisfied, andfis continuous atx=0 . Thus,fis continuous forall values ofx .



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ii.) .

Since

iii.) ,

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

all values ofx . Now show thatfis differentiable for all values ofx . For we

can differentiatefusing the product rule and the chain rule. That is, for thederivative offis

.

Use the limit definition of the derivative to differentiatefatx=0 . Then


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.

Use the Squeeze Principle to evaluate this limit. For

.

If , then

.

If , then

.

In either case,

,

and it follows from the Squeeze Principle that

.

Thus,fis differentiable for all values ofx . Check to see iff' is continuous atx=0 .The functionf' is defined atx=0 since

i.)f'(0) = 0 .

However,

ii.)


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does not exist since the values of oscillate between -1 and +1asx approaches zero. Thus, condition ii.) is violated, and the derivative ,f' , is notcontinuous atx=0 .

NOTE : The continuity of functionffor all values ofx also follows from the factthatfis differentiable for all values ofx .

LIMITES DE FUNES VIA O PRINCPIO DO SANDUCHE

Os seguintes problemas envolvem o clculo de limites usando o princpio do

sanduche que est dado abaixo:

SQUEEZE PRINCIPLE : Suponha que as funesf, g , e h satisfaam

e

.

Ento

.

(NOTA : A quantidade A pode ser um nmero finito , , or . A quantidade L

pode ser um nmero finito, , or . O princpio do sanduche usado emproblemas de limites onde os procedimentos algbricos (fatorao, conjugao emanipulao algbrica etc.) no so efetivos. O uso do princpio do sanduche requeranlise acurada, habilidade algbrica e uso cuidadoso de desigualdades.


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.


.


.


.


.


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.

.


.

.

o PROBLEMA 8 : Assuma que existe

e . Encontre .

.

o PROBLEMA 9 : Considere um crculo de raio 1 centrado na origem e um ngulo

de radians, , no diagrama.


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a.) Considerando as reas dos tringulo reto setor OAD, setor OAC, e tringuloreto OBC, conclua que

.

b.) Use parte a.) e o princpio do sanduche para mostrar que

o PROBLEMA 10 : Suponha que


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Mostre quef contnua emx=0

SOLUTIONS TO LIMITS USING THE SQUEEZE PRINCIPLE

SOLUTION 1 : First note that

because of the well-known properties of the sine function. Since we are computing the limitasx goes to infinity, it is reasonable to assume thatx > 0 . Thus,

.

Since

,

it follows from the Squeeze Principle that

.


http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%201http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%201http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%201http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%201


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because of the well-known properties of the cosine function. Now multiply by -1, reversingthe 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 thatx + 3 >0. Thus,

.

Since

,


.




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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 that 3 -2x < 0. Now divide each component by 3 - 2x, reversing the inequalities and getting

,

or

.

Since

,


.


SOLUTION 4 : Note that DOES NOT EXIST since values

of oscillate between -1 and +1 asx approaches 0 from the left. However, this does


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NOT necessarily mean that does not exist ! ? #. Indeed,x3 < 0 and

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

or

.

Since

,


.



,

so that


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= (does not exist).



,

so that

,

,

and

.

Then

=

=


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or

.

Now divide byx2 + 1 and multiply byx2 , getting

.

Then

=

=

=

=

= 0 .


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Similarly,

= 0 .

It follows from the Squeeze Principle that

= 0 .


SOLUTION 8 : Since

=

and

= ,


,

that is,


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.

Thus,

.


SOLUTION 9 : a.) First note that (See diagram below.)

area of triangle OAD < area of sector OAC < area of triangle OBC .


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The area of triangle OAD is

(base) (height) .

The area of sector OAC is

(area of circle) .

The area of triangle OBC is

(base) (height) .

It follows that


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or

.

b.) If , then and , so that dividing by results in

.

Taking reciprocals of these positive quantities gives

or

.

Since

,


.



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SOLUTION 10 : Recall that functionfis continuous atx=0 if

i.)f(0) is defined ,

ii.) exists ,

and

iii.) .

First note that it is given that

i.)f(0) = 0 .

Use the Squeeze Principle to compute . For we know that

,

so that

.

Since


ii.) .

Finally,

iii.) ,


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resolução limites

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