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I cannot seem to figure out why the answer to this question is 1/12 $$ \lim_{x \to 2} \frac{x-2}{x^3-8}. $$ I can't seem to find a way to factor it so I keep getting $0/0$. This is from a text book and the text book answer key states the answer is $1/12$.

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    $\begingroup$ $8=2^3$, and therefore $x^3-2^3=(x-2)(x^2+2x+4)$ $\endgroup$
    – Siminore
    Mar 8, 2014 at 12:59
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    $\begingroup$ Use this: $x^3-8=x^3-2^3=(x-2)(x^2+2x+4)$ $\endgroup$
    – Mher
    Mar 8, 2014 at 12:59
  • $\begingroup$ Google "factor difference of cubes". $\endgroup$ Mar 8, 2014 at 13:10

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$$ x^3-8 = (x-2)(x^2+2x+4) \\ \lim_{x \rightarrow 2} \frac{(x-2)}{(x^3-8)} \\ \lim_{x \rightarrow 2} \frac{(x-2)}{(x-2)(x^2+2x+4)} \\ \lim_{x \rightarrow 2} \frac{1}{(x^2+2x+4)} = \frac{1}{12}\\ $$

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Well you can use Bezouts little theorem,the remainder of dividing the polynomial $P(x)$ with $(x-a)$ is equal to $P(a)$,so basically if $P(x)=x^3-8$ then the remainder of dividing it with $x-2$ is equal to $P(2)=2^3-8=0$,which means $(x-2)\mid P(x)$,you can either divide the polynomial or use the known formula $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$

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