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Evaluate $\lim_{x \to 2} \frac{|x^3 - 8|}{x - 2}$

How do I evaluate this limit? I get an intermediate form of $\frac{0}{0}$ and couldn't seem to do anything algebraically testing the limit from the positive and negative side as I wasn't entirely sure how to treat the absolute value term. Should I just try to plot some points and observe what is happening around $x$ = 2?

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Hint. $x^3 - 8 = (x-2)(x^2+2x+4)$; then handle the cases $x\to 2^+$ and $x\to 2^-$ separately.

Note for future: If you have a polynomial $p(x)$, and $p(a)=0$ for some number $a$, then you can always factor out $x-a$; that is, you can always write $p(x)$ as $(x-a)(\text{something})$.

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    @stariz77: Right: you cannot simply cancel through the absolute value, because you don't know whether $|x-2|=x-2$ or $|x-2|=2-x$. That's why you handle the case $x\to 2^-$ and the case $x\to 2^+$ separately. The absolute value is often a signal that you should try one-sided limits.2012-03-23