### 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.)