Limits of Functions As X Approaches Infinity

LIMITS OF FUNCTIONS AS X APPROACHES INFINITY

The following problems require the algebraic computation of limits of functions as x approaches plus or minus infinity. 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 forms and during the computations of these limits. Initially, many students INCORRECTLY conclude that is equal to 1 , or that the limit does not exist, or is or . Many also conclude that is equal to 0 . In fact, the forms and are examples of indeterminate forms. This simply means that you have not yet determined an answer. Usually, these indeterminate forms can be circumvented by using algebraic manipulation. Such tools as algebraic simplification and conjugates can easily be used to circumvent the forms and so that the limit can be calculated.

SOLUTIONS TO LIMITS OF FUNCTIONS AS X APPROACHES PLUS OR MINUS INFINITY

SOLUTION 1 :

= = 0 .

(The numerator is always 100 and the denominator approaches as x approaches , so that the resulting fraction approaches 0.)

SOLUTION 2 :

= = 0 .

(The numerator is always 7 and the denominator approaches as x approaches , so that the resulting fraction approaches 0.)

SOLUTION 3 :

(This is NOT equal to 0. It is an indeterminate form. It can be circumvented by factoring.)

(As x approaches , each of the two expressions and 3 x - 1000 approaches .)

(This is NOT an indeterminate form. It has meaning.)

= .

(Thus, the limit does not exist. Note that an alternate solution follows by first factoring out , the highest power of x . Try it.)

SOLUTION 4 :

(As x approaches , each of the two expressions and approaches . )

(This is NOT an indeterminate form. It has meaning.)

= .

(Thus, the limit does not exist. Note that an alternate solution follows by first factoring out , the highest power of x . Try it.)

SOLUTION 5 :

(Note that the expression leads to the indeterminate form . Circumvent this by appropriate factoring.)

= .

(As x approaches , each of the three expressions , , and x - 10 approaches .)

= .

(Thus, the limit does not exist. Note that an alternate solution follows by first factoring out , the highest power of x . Try it. )

SOLUTION 6 :

(This is an indeterminate form. Circumvent it by dividing each term by x .)

(As x approaches , each of the two expressions and approaches 0.)

= .

SOLUTION 7 :

(Note that the expression leads to the indeterminate form as x approaches . Circumvent this by dividing each of the terms in the original problem by .)

(Each of the three expressions , , and approaches 0 as x approaches .)

= .

SOLUTION 8 :

(Note that the expression leads to the indeterminate form as x approaches . Circumvent this by dividing each of the terms in the original problem by , the highest power of x in the problem . This is not the only step that will work here. Dividing by , the highest power of x in the numerator, also leads to the correct answer. You might want to try it both ways to convince yourself of this.)

(Each of the five expressions , , , , and approaches 0 as x approaches .)

= 0 .

SOLUTION 9 :

(Note that the expression leads to the indeterminate form as x approaches . Circumvent this by dividing each of the terms in the original problem by , the highest power of x in the problem. . This is not the only step that will work here. Dividing by x , the highest power of x in the denominator, actually leads more easily to the correct answer. You might want to try it both ways to convince yourself of this.)

(Each of the three expressions , , and approaches 0 as x approaches .)

(This is NOT an indeterminate form. It has meaning. However, to determine it's exact meaning requires a bit more analysis of the origin of the 0 in the denominator. Note that = . It follows that if x is a negative number then both of the expressions and are negative so that is positive. Thus, for the expression the numerator approaches 7 and the denominator is a positive quantity approaching 0 as x approaches . The resulting limit is .)

= .

(Thus, the limit does not exist.)

SOLUTION 10 :

(You will learn later that the previous step is valid because of the continuity of the square root function.)

(Inside the square root sign lies an indeterminate form. Circumvent it by dividing each term by , the highest power of x inside the square root sign.)

(Each of the two expressions and approaches 0 as x approaches .)

= .

SOLUTION 11 :

= `` ''

(Circumvent this indeterminate form by using the conjugate of the expression in an appropriate fashion.)

(Recall that .)

= 0 .

SOLUTION 12 :

(This is NOT an indeterminate form. It has meaning.)

= .

(Thus, the limit does not exist.)