Question

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

Answer 1

We have,

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

This is $\frac{0}{0}$ form.

So, apply L-hospital rule,

$\underset{x\to 2}{lim}\frac{3x^2+6x-9-0}{3x^2-1-0}$

$\underset{x\to 2}{lim}\frac{3x^2+6x-9}{3x^2-1}$

$=\frac{3\times 2^2+6\times 2-9}{3\times 2^2-1}$

$=\frac{3\times 4+6\times 2-9}{3\times 4-1}$

$=\frac{12+12-9}{12-1}$

$=\frac{15}{11}$

Hence, this is the answer.