Solutions to Limits as x Approaches a Constant

### SOLUTIONS TO LIMITS OF FUNCTIONS AS X APPROACHES A CONSTANT

SOLUTION 1 :

.

SOLUTION 2 :

(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. )

SOLUTION 3 :

(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. )

.

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

.

(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. )

.

(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 .

(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. )

.

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

(Use the well-known fact that .)

.

(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. )

.

(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.)

(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.)

(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.

(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 .

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

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.) .

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

Thus,

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

iff b-12+ 8b = 3

iff 9b = 15

iff .

Then

.