Question 1020220
{{{lim(x->-1, ((ax+abs(x+1))*abs(x+b-2))/abs(x+1))}}}

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(x+1)*abs(x+b-2)}}}
==> the quotient becomes {{{abs(x+b-2)}}} after division by {{{abs(x+1)}}}, hence any value of b will render the limit existent.

If {{{a <> 0}}}, then the numerator {{{(ax+abs(x+1))*abs(x+b-2)}}} approaches {{{-a*abs(b-3)}}} as x approaches -1, and the only possibility is for b = 3.  The quotient then becomes 
{{{((ax+abs(x+1))*abs(x+1))/abs(x+1) = ax+abs(x+1)}}} after division, and the limit of this expression exists as x -> -1 for whatever value of a.

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.