# Finding $\lim\limits_{x\rightarrow 2} \dfrac{x^3-8}{x^2-x-2}$

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.

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The denominator factors as $x^2-x-2 = (x-2)(x+1)$. Does that help? –  Arturo Magidin Oct 8 '11 at 20:35
If you plug $2$ into a polynomial and get $0$, then $x-2$ is one of its factors. That's worth knowing. And if you plug $2$ in and get something other than $0$, then you won't get a $0$ in the denominator in a case like this, so then you could just plug $2$ into the whole expression and that's the limit. –  Michael Hardy Oct 8 '11 at 21:20

Hint: Note that $x^2-x-2=(x-2)(x+1)$.

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And theres my answer.. I didn't even think that the denominator was factorable! How could I have missed that. Thank you - I will approve your answer when the time limit for which I can't approve any answer is up. –  Mike Gates Oct 8 '11 at 20:36
@Mike: If $p(x)$ is a polynomial, and $p(a)=0$, then $p(x)$ can always be factored as $p(x)=(x-a)q(x)$ for some polynomial $q(x)$. This is the Factor Theorem, so with these kinds of limits (a rational function that evaluates to $\frac{0}{0}$), you can always factor both the numerator and the denominator, cancel the factor, and try again. –  Arturo Magidin Oct 8 '11 at 20:37
Oh okay. That's sensible. –  Mike Gates Oct 8 '11 at 20:44

$$\lim_{x \to 2}\dfrac{x^3-8}{x^2-x-2}=\lim_{x \to 2}\dfrac{(x^3-8)'}{(x^2-x-2)'}=...$$

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You can use the L`Hospital rule to find the answer.

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Why this late answer to a question with an accepted answer? And it's not like your answer is much better than the previous answers either. –  TMM Mar 30 '13 at 15:42