SOLUTION 1 :
=
= 0 .
(The numerator is always 100 and the denominator approaches as x approaches , so that the resulting fraction approaches 0.)
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SOLUTION 2 :
=
= 0 .
(The numerator is always 7 and the denominator approaches as x approaches , so that the resulting fraction approaches 0.)
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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 - 100 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.)
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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.)
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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. )
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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.)
=
= .
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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 .)
=
= .
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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 .
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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.)
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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 .)
=
=
= .
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SOLUTION 11 :
= `` ''
(Circumvent this indeterminate form by using the conjugate of the expression in an
appropriate fashion.)
=
(Recall that .)
=
=
=
=
= 0 .
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SOLUTION 12 :
=
(This is NOT an indeterminate form. It has meaning.)
= .
(Thus, the limit does not exist.)
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