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

SOLUTION 13 :

=

(This is true because the expression approaches and the expression x + 3 approaches as x approaches . The next step follows from the following simple fact. If A is a positive quantity, then = A . )

=

=

=

(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 three expressions , , and approaches 0 as x approaches .)

=

=

= .

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SOLUTION 14 :

=

(This is true because the expression approaches and the expression x + 3 approaches as x approaches . The next step follows from the following simple fact. If A is a negative quantity, then = - A so that = - ( - A ) = A . Please make sure that you think about and understand this before proceeding. )

=

=

=

(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 three expressions , , and approaches 0 as x approaches .)

=

=

= .

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SOLUTION 15 :

=

(You will learn later that the previous step is valid because of the continuity of the logarithm function. Note also that the expression leads to the indeterminate form . Circumvent it by dividing each term by , the highest power of x .)

=

=

=

(The term approaches 0 as x approaches .)

=

=

= 0 .

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SOLUTION 16 :

=

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

=

=

(The expression leads to the indeterminate form . Circumvent it by dividing each term by , the highest power of x in the expression.)

=

=

=

(Each of the terms and approaches 0 as x approaches .)

=

=

= .

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SOLUTION 17 :

(As x approaches each of the expressions and approaches 0. The following steps explain why.)

=

=

=

=

= 0 .

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SOLUTION 18 :

=

(Circumvent this indeterminate form by dividing each term in the expression by . Division by also works . You might want to try it both ways to convince yourself of this. Also, BEWARE of making one of the following common MISTAKES : = or \ = .)

=

=

=

(Since approaches 0 and approaches as x approaches , we get the following resultant limit.)

=

= .

(Thus, the limit does not exist.)

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SOLUTION 19 :

= `` '' truein truein (BEWARE of making the following common MISTAKE : = . Realize also that the form `` '' is an indeterminate one ! It is not equal to 1 ! Circumvent it in the following algebraic ways.)

=

=

(Factor out the term . If you have time, try factoring out the term to convince yourself that it DOESN'T seem to help !)

=

=

=

=

=

(The expressions and approach 0 as x approaches .)

=

= .

= 9 .

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Duane Kouba
Wed Apr 2 10:10:40 PST 1997