SOLUTION: For what value of the constants a and b such that the following limit exists: lim x→−1 (ax + |x + 1|)|x + b − 2|/ |x + 1| .

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Question 1020220: For what value of the constants a and b such that the following limit exists:
lim
x→−1
(ax + |x + 1|)|x + b − 2|/
|x + 1|
.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
lim%28x-%3E-1%2C+%28%28ax%2Babs%28x%2B1%29%29%2Aabs%28x%2Bb-2%29%29%2Fabs%28x%2B1%29%29
Since the denominator approaches 0 as x goes to 0, the numerator also needs to go to zero.
If a = 0, then the numerator becomes abs%28x%2B1%29%2Aabs%28x%2Bb-2%29
==> the quotient becomes abs%28x%2Bb-2%29 after division by abs%28x%2B1%29, hence any value of b will render the limit existent.
If a+%3C%3E+0, then the numerator %28ax%2Babs%28x%2B1%29%29%2Aabs%28x%2Bb-2%29 approaches -a%2Aabs%28b-3%29 as x approaches -1, and the only possibility is for b = 3. The quotient then becomes
%28%28ax%2Babs%28x%2B1%29%29%2Aabs%28x%2B1%29%29%2Fabs%28x%2B1%29+=+ax%2Babs%28x%2B1%29 after division, and the limit of this expression exists as x -> -1 for whatever value of a.
Thus {a, b epsilonR|a = 0 and b is any real number} U { a,b epsilon R| b=3 and a is any real number} would render the limit above existent.