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