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

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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1 Answer

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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Ah! you beat my hint for like two seconds. Im going to delete mine :P – Daniel Montealegre Mar 23 '12 at 5:01
Can I cancel the (x-2) term through the absolute value signs? Assuming I can, the $x^2 + 2x + 4$ term is always positive, so the absolute value signs are then gone right? and I can evaluate the limit of the remaining term. – stariz77 Mar 23 '12 at 5:09
Actually, looking at your answer again, I test both cases $\pm (x^3 - 8)$. I assume the negative case is the $x\rightarrow 2^{-}$ and the positive case is $x\rightarrow 2^{+}$ – stariz77 Mar 23 '12 at 5:14
@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. – Arturo Magidin Mar 23 '12 at 5:17