.

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

(Divide out the factors *x* - 2 , the factors which are causing the indeterminate form . Now the limit can be
computed. )

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

(Divide out the factors *x* - 3 , the factors which are causing the indeterminate form . Now the limit can be
computed. )

.

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(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 factors *x* + 2 , the factors which are causing the indeterminate form . Now the limit can be
computed. )

.

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

. )

(Divide out the factors *x* - 4 , the factors which are causing the indeterminate form . Now the limit can be
computed. )

.

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(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, write *x* - 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. )

= 27 .

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(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.)

(Divide out the factors *x* - 1 , the factors which are causing the indeterminate form . Now the limit can be
computed. )

.

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

(Use the well-known fact that .)

.

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(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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(Factor *x* 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.)

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(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.)

.

(Thus, the limit does not exist.)

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(Recall that . )

(Divide out the factors *x* - 1 , the factors which are causing the indeterminate form . Now the limit
can be computed. )

.

(The numerator approaches 3 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. *

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(Make the replacement so that . Note that as
*x* 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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* SOLUTION 14 :* Consider the function

i.) The graph of *f* is given below.

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.) .
- k.) .
- l.) .

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* SOLUTION 15 :* Consider the function

Determine the values of constants *a* and *b* so that
exists. Begin by computing one-sided limits at *x*=2 and setting each equal to 3. Thus,

and

.

Now solve the system of equations

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

Thus,

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

iff *b*-12+ 8*b* = 3

iff 9*b* = 15

iff .

Then

.

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Tue Aug 27 13:48:42 PDT 1996