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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