Question

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

Answer 1

We have,

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

This is $\frac{0}{0}$ form, so apply L-Hospital rule

$\underset{x\to \sqrt{2}}{lim}\frac{2x-0}{2x+3\sqrt{2}-0}$

$\underset{x\to \sqrt{2}}{lim}\frac{2x}{2x+3\sqrt{2}}$

$=\frac{2\sqrt{2}}{2\sqrt{2}+3\sqrt{2}}$

$=\frac{2\sqrt{2}}{5\sqrt{2}}$

$=\frac{2}{5}$

Hence, this is the answer.