Limit of (x^2-4x+4)/(x^2+x-6) as x approaches 2

(x-2)^2
------------
(x-2)(x+3)
x-2
= --------
x+3
as x ->2
we get 0

or by hopital's theory, differentiating the nominator and the denominator, we get: 2x-4/2x+1
then by substituting, 0/5=zero

Factor
(x-2)(x-2) / (x+3)(x-2)
The x-2 cancels
(x-2) / (x+3)
Then sub in the 2 for the x
(2-2) / (2+5)
0/5
0

So the answer is 0

this simplifies to:

(x-2)(x-2)
-------------
(x-2)(x+3)

which simplifies to:

(x-2)
------
(x+3)

So, x approaches 2, therefore it = 0