I've done so many limit problems in calculus lately, but I can't wrap my mind around how to simplify this one in order to solve it:
$$
\lim_{x\rightarrow 2} \dfrac{x^3-8}{x^2-x-2}
$$
I understand the $x^3-8$ factors down to $(x-2)(x^2+2x+4)$, but that still leaves us with
$$
\lim_{x\rightarrow 2} \dfrac{(x-2)(x^2+2x+4)}{x^2-x-2},
$$
which I can't seem to find a way to simplify so that the denominator is not equal to 0.
In case anyone figures out themselves, the answer is 4 (I was given the answer - this is on a review sheet for an upcoming exam). Also, I tagged this as homework, even though it is not technically homework.
So if anyone could help point me in the right direction here, that would be very helpful.
Answer
Hint: Note that $x^2-x-2=(x-2)(x+1)$.
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