Solutions to Limits of Functions as x Approaches a Constant

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SOLUTIONS TO LIMITS OF FUNCTIONS AS X APPROACHES A

CONSTANT

SOLUTION 1:

.

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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. )

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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 . Now the 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 . Now the 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

. )

(Divide out the factorsx - 4 , the factors which are causing the indeterminate

form . Now the limit can be computed. )

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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.)

(Divide out the factorsx - 1 , the factors which are causing the indeterminate

form . Now the limit can be computed. )

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.

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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. )

.

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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.)

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SOLUTION 11 :

(The numerator approaches -3 and the denominator is a negative quantity whichapproaches 0 as x

approaches 0 .)

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

.

(Thus, the limit does not exist.)

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SOLUTION 12 :

(Recall that . )

(Divide out the factorsx - 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.)

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The limit does not exist.

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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 .

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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.

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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 numeratoris 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 .

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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 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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.

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About this document ...

Duane Kouba

Tue Aug 27 13:48:42 PDT 1996