resoluçao lista calculo 28

8/2/2019 resoluao lista calculo 28

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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 the limit can be computed. )


SOLUTION 3 :

(Circumvent the indeterminate form by factoring both the numerator and denominator.)
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%201http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%202http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%202http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%201


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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 .)
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%203http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%203


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(Factor the denominator . Recall that .)

(Divide out the factorsx + 2 , the factors which are causing the indeterminate form .


.


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

reasonable next step.

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

recall that

.)


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(Divide out the factors , the factors which are causing the indeterminate form

. Now the limit can be computed. )

= 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 next few 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 .)

.


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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. )

.


SOLUTION 10 :


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(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 as x

approaches 0 .)

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

.


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(Thus, the limit does not exist.)


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 is NEITHER

NOR . The correct answer follows.)

The limit does not exist.


SOLUTION 13 :


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(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 . )

(Recall that . )

= 2 .



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

SOLUTION 14 : Consider the function

i.) The graph offis given below.

ii.) Determine the following limits.

a.) .

b.)

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

.

d.) .

e.) .


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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.) .

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


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iff .

Then

.

resoluçao lista calculo 28

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