SOLUTION 1 :
.
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SOLUTION 2 :
(Circumvent the indeterminate form by factoring both the numerator and denominator.)
(Divide out the factors x - 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.)
(Divide out the factors x - 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 .)
(Factor the denominator . Recall that
.)
(Divide out the factors x + 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
. )
(Divide out the factors x - 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, write x - 27 as the difference of cubes and recall that
.)
(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.)
(Divide out the factors x - 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 :
(Factor x 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.)
.
(Thus, the limit does not exist.)
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SOLUTION 12 :
(Recall that . )
(Divide out the factors x - 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 :
(Make the replacement so that . Note that as
x 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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The next problem requires an understanding of one-sided limits.
SOLUTION 14 : Consider the function
i.) The graph of f is given below.
ii.) Determine the following limits.
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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 at x=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
iff b-12+ 8b = 3
iff 9b = 15
iff .
Then
.
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