Question

Find the following limit.
$\underset{x\to 2}{lim}\frac{x^3-3x-2}{x^3-8}.$

Answer 1

$\underset{x\to 2}{lim}\frac{x^3-3x-2}{x^3-8}$

By using factorization $[x-a^3-b^3=(a-b\left)\right(a^2+ab+b^2\left)\right]$

$\underset{x\to 2}{lim}\frac{(x-2)(x^2+2x+1)}{(x-2)(x^2+2x+4)}$

$\underset{x\to 2}{lim}\frac{x^2+2x+1}{x^2+2x+4}$

$=\underset{x\to 2}{lim}\frac{x^2+2x+1}{x^2+2x+4}$

$=\frac{4+4+1}{4+4+4}=\frac{9}{12}=\frac{3}{4}$.