Math Q:Limits and continuous

g(x) = (x - 1)2(x + 2)/(x - A)

So when A = 1:

g(x) = (x - 1)(x + 2) when x =/= 1

lim (x → 1) g(x) = 0

Hence g(x) = 0 when x = 1 for it to be continuous at x = 1

So B = 1

When A = -2:

g(x) = (x - 1)2 when x =/= -2

lim (x → -2) g(x) = 9

Hence g(x) = 9 when x = -2 for it to be continuous at x = -2

So B = 9

Therefore there are 2 sets of A and B for g(x) to be continuous for all real x:

(i) A = 1 and B = 1

(ii) A = -2 and B = 9

2011-06-16 11:16:30 補充：
Yes, x^3 -3x +2 = (x - 1)^2 (x + 2) exactly