Question

Evaluate the limit:
$\underset{x\to 2}{lim}\frac{x^3+3x^2-9x-2}{x^3-x-6}$

Answer 1

Given: $\underset{x\to 2}{lim}\frac{x^3+3x^2-9x-2}{x^3-x-6}$ becomes $\frac{0}{0}$ form

$\Rightarrow x^3+3x^2-9x-2=(x-2)(x^2-5x+1)$

Also,

$\Rightarrow x^3-x-6=(x-2)(x^2+2x+3)$

$\therefore \underset{x\to 2}{lim}\frac{x^3+3x^2-9x-2}{x^3-x-6}$

$=\underset{x\to 2}{lim}\frac{(x-2)(x^2+5x+1)}{(x-2)(x^2+2x+3)}$

$\Rightarrow \underset{x\to 2}{lim}\frac{(x^2+5x+1)}{(x^2+2x+3)}$

$=\frac{(2^2+5(2)+1)}{(2^2+2(2)+3)}$

$=\frac{15}{11}$

$Therefore,$

$\underset{x\to 2}{lim}\frac{x^3+3x^2-9x-2}{x^3-x-6}=\frac{15}{11}$