# Thread: Really Simple Limit Question

1. ## Really Simple Limit Question

I was just wandering if we had say $\displaystyle \frac{(x^2 - x -6)^2}{(x+2)^2}$, if we had to find the limit of that as it approached -2 for example.
Could I cancel both top and bottom by (x + 2)^2, or would that not be correct since (x+2) = 0 when x = -2

2. ummm im not surei know u could write it as $\displaystyle \frac{(x - 3)^2(x + 2)^2}{(x + 2)^2}$ but im not sure if ur allowed to cancel since the bottom tends to 0.

3. Hello,
Originally Posted by Lipticboven
ummm im not surei know u could write it as $\displaystyle \frac{(x - 3)^2(x + 2)^2}{(x + 2)^2}$ but im not sure if ur allowed to cancel since the bottom tends to 0.
You can simplify, and that's the principle of these limits.

$\displaystyle x^2-x-6=(x-3)(x+2)$

So, firstly without working with limits, $\displaystyle \frac{(x^2-x-6)^2}{(x+2)^2}=\frac{(x-3)^2(x+2)^2}{(x+2)^2}=(x-3)^2$
You can simplify, because you don't know yet top and bottom are 0. There's no problem.

Now, take the limit as x tends to 2.