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SOLUTION 1 :
.
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SOLUTION 2 :
(Circumvent the indeterminate form by factoring both the numerator and denominator.)
(Divide out the factorsx - 2 , the factors which are causing the indeterminate form .
Now the limit can be computed. )
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SOLUTION 3 :
(Circumvent the indeterminate form by factoring both the numerator and denominator.)
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%201http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%202http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%202http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%201 -
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(Divide out the factorsx - 3 , the factors which are causing the indeterminate form .
Now the limit can be computed. )
.
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SOLUTION 4 :
(Algebraically simplify the fractions in the numerator using a common denominator.)
(Division by is the same as multiplication by .)
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%203http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%203 -
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(Factor the denominator . Recall that .)
(Divide out the factorsx + 2 , the factors which are causing the indeterminate form .
Now the limit can be computed. )
.
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SOLUTION 5 :
(Eliminate the square root term by multiplying by the conjugate of the numerator over
itself. Recall that
. )
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%204http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%204 -
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(Divide out the factorsx - 4 , the factors which are causing the indeterminate form .
Now the limit can be computed. )
.
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SOLUTION 6 :
(It may appear that multiplying by the conjugate of the numerator over itself is a
reasonable next step.
It's a good idea, but doesn't work. Instead, writex - 27 as the difference of cubes and
recall that
.)
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%205http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%205 -
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(Divide out the factors , the factors which are causing the indeterminate form
. Now the limit can be computed. )
= 27 .
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SOLUTION 7 :
(Multiplying by conjugates won't work for this challenging problem. Instead, recall that
and ,
and note that and . This should help
explain the next few mysterious steps.)
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(Divide out the factorsx - 1 , the factors which are causing the indeterminate form .
Now the limit can be computed. )
.
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SOLUTION 8 :
(If you wrote that , you are incorrect. Instead, multiply and divide by
5.)
(Use the well-known fact that .)
.
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SOLUTION 9 :
(Recall the trigonometry identity .)
(The numerator is the difference of squares. Factor it.)
(Divide out the factors , the factors which are causing the indeterminate form
. Now the limit can be computed. )
.
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SOLUTION 10 :
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(Factorx from the numerator and denominator, then divide these factors out.)
(The numerator approaches -7 and the denominator is a positve quantity approaching 0 .)
(This is NOT an indeterminate form. The answer follows.)
.
(Thus, the limit does not exist.)
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SOLUTION 11 :
(The numerator approaches -3 and the denominator is a negative quantity which
approaches 0 as x
approaches 0 .)
(This is NOT an indeterminate form. The answer follows.)
.
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(Thus, the limit does not exist.)
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SOLUTION 12 :
(Recall that . )
(Divide out the factorsx - 1 , the factors which are causing the indeterminate form .
Now the limit can be computed. )
.
(The numerator approaches 1 and the denominator approaches 0 as x approaches 1 .
However, the quantity
in the denominator is sometimes negative and sometimes positive. Thus, the correct
answer is NEITHER
NOR . The correct answer follows.)
The limit does not exist.
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SOLUTION 13 :
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2011http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2012http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2011http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2012 -
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(Make the replacement so that . Note that asx approaches , h
approaches 0 . )
(Recall the well-known, but seldom-used, trigonometry identity
.)
(Recall the well-known trigonometry identity . )
(Recall that . )
= 2 .
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http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2013http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2013 -
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The next problem requires an understanding of one-sided limits.
SOLUTION 14 : Consider the function
i.) The graph offis given below.
ii.) Determine the following limits.
a.) .
b.)
c.) We have that does not exist since does not equal
.
d.) .
e.) .
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f.) We have that since .
g.) We have that (The
numerator is always -1 and the denominator is always a positive number
approaching 0.) , so the limit does not exist.
h.) .
i.) We have that does not exist since does not equal
.
j.) .
k.) .
l.) .
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SOLUTION 15 : Consider the function
Determine the values of constants a and b so that exists. Begin by computing
one-sided limits atx=2 and setting each equal to 3. Thus,
and
.
Now solve the system of equations
a+2b = 3 and b-4a = 3 .
Thus,
a = 3-2b so that b-4(3-2b) = 3
iffb-12+ 8b = 3
iff 9b = 15
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iff .
Then
.