SOLUTION: f(x) = (ax^2 + a^2(x) - 2)/(x^3 - 3x + 2) is there a real number of a in which lim x -> 1 f(x) exists? If so what is the value of a and what is the limit?

Factor the denominator using potential zeros and synthetic division.

Potential zeros are ±1, ±2.  Try 1

1|1 0 -3  2
 |  1  1 -2
  1 1 -2  0

So it works and thus we have factored the denominator as

(x-1)(x²+x-2)
(x-1)(x-1)(x+2)
(x-1)²(x+2)



The factor (x+2) does not become 0 when x=1, so if the
numerator could be a multiple of (x-1)² then that factor
would cancel into the top, creating a new function g(x)
which is equal to f(x) for all x except 1.  Its value
at 1 would be defined, and thus f(x) would approach its
value as x->1. So we will see if there is a constant k 
that would permit this cancellation. If so, we would
have this identity:







So the constant terms would have to be the same:

k=-2.  And if we substitute k=-2:




So they will be identical if a = -2. 




So we substitute for the numerator in



and get



Now, after canceling, we have the new function 
g(x) which is identical to f(x) except for one
point:



Then the limit of f(x) as x -> 1 will equal g(1) 
which is



Answer: a=-2 and the limit = -2/3.

Edwin