document.write( "Question 1020220: For what value of the constants a and b such that the following limit exists:
\n" ); document.write( "lim
\n" ); document.write( "x→−1
\n" ); document.write( "(ax + |x + 1|)|x + b − 2|/
\n" ); document.write( "|x + 1|
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Algebra.Com's Answer #636140 by robertb(5830)\"\" \"About 
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\"lim%28x-%3E-1%2C+%28%28ax%2Babs%28x%2B1%29%29%2Aabs%28x%2Bb-2%29%29%2Fabs%28x%2B1%29%29\"\r
\n" ); document.write( "\n" ); document.write( "Since the denominator approaches 0 as x goes to 0, the numerator also needs to go to zero. \r
\n" ); document.write( "\n" ); document.write( "If a = 0, then the numerator becomes \"abs%28x%2B1%29%2Aabs%28x%2Bb-2%29\"
\n" ); document.write( "==> 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.\r
\n" ); document.write( "\n" ); document.write( "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
\n" ); document.write( "\"%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.\r
\n" ); document.write( "\n" ); document.write( "Thus {a, b \"epsilon\"R|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. \n" ); document.write( "
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