# How to solve Limit, can't seem to factor.

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 $1/12$.

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

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