explain why this is correct lim -> 2...?

Question

lim x-->2 of "(x^2)+ x - 6/ x -2" = lim x --> 2 of "(x+3)"

I don't see how this is possibly correct

Answer 1

hi ar j.

this works because you can factor the numerator (x^2+x-6) into (x+3) times (x-2). You end up with [(x+3)*(x-2)] all divided by (x-2). you now have two (x-2)'s, one in num. and one in denom., so they cancel out... this leaves you with lim x->2 of (x+3).

There ya go =D

Answer 2

(x^2+x-6)/(x-2) = (x+3)(x-2)/(x-2)
for x != 2, this = x+3. So, as x approaches 2, f(x) approaches 5