Limits of Functions As X Approaches a Constant

LIMITS OF FUNCTIONS AS X APPROACHES A CONSTANT

The following problems require the use of the algebraic computation of limits of functions as x approaches a constant. Most problems are average. A few are somewhat challenging. All of the solutions are given WITHOUT the use of L'Hopital's Rule. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by giving careful consideration to the form during the computations of these limits. Initially, many students INCORRECTLY conclude that is equal to 1 or 0 , or that the limit does not exist or is or . In fact, the form is an example of an indeterminate form. This simply means that you have not yet determined an answer. Usually, this indeterminate form can be circumvented by using algebraic manipulation. Such tools as algebraic simplification, factoring, and conjugates can easily be used to circumvent the form so that the limit can be calculated.

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. )

.

SOLUTION 5 :

(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. )

.

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, 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 .

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

.

SOLUTION 8 :

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

(Use the well-known fact that .)

.

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 :

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

(The numerator approaches -7 and the denominator is a positive 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.)

.

(Thus, the limit does not exist.)

SOLUTION 12 :

(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.

SOLUTION 13 :

(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

.

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