<s>[INST] How to find the following limit algebraically? I've been trying to answer this for a while and I know it's a simple question relative to most questions that are posted here.
$$
\lim_{x\rightarrow -2}\: \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)}
$$
If we substitute -2 for $ x $ we get $ 0/0 $, an indeterminate form. I figured that the denominator can be rewritten as $ (x^2-4)(x+1) $. And then I tried to factor something in the numerator but couldn't see anything interesting. How do I find the limit algebraically?
I know that the answer is supposed to be 1, but I don't know how they got there.
Thanks in advance!
 [/INST] Notice that:
$$x^4+5x^3+6x^2 = x^2(x^2+5x+6) = x^2(x+2)(x+3)$$
and
$$x^2(x+1)-4(x+1) = (x^2-4)(x+1) = (x+2)(x-2)(x+1).$$
Then, you have:
$$\lim_{x\rightarrow -2}\: \frac{x^2(x+2)(x+3)}{(x+2)(x-2)(x+1)} = \\
 = \lim_{x\rightarrow -2}\: \frac{x^2(x+3)}{(x-2)(x+1)} = \\
= \frac{(-2)^2(-2+3)}{(-2-2)(-2+1)} = \frac{4}{4} = 1.$$
</s>