Question

Calculate the following limits.
$\underset{x\to 2}{lim}\frac{\sqrt{x^2-1}-\sqrt{x^2+x3}}{x^2-4}.$

Answer 1

$\underset{x\to 2}{lim}\frac{\sqrt{x^2-1}-\sqrt{x^2+x-3}}{x^2-4}\left[\frac{0}{0}form\right]$

Apply L'hospital rule

$\underset{x\to 2}{lim}\left[\frac{\frac{x}{\sqrt{x^2-1}}-\frac{2x+1}{2\sqrt{x^2+x-3}}}{2x}\right]$

$\Rightarrow \underset{x\to 2}{lim}\left[\frac{2x\sqrt{x^2+x-3}-(2x+1)\sqrt{x^2-1}}{4x\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2+x-3}\right)}\right]$

$\Rightarrow$ Put the limit

$\Rightarrow \frac{2\left(2\right)\sqrt{6+(-3)}-\left(5\right)\sqrt{3}}{8\sqrt{3}\left(\sqrt{3}\right)}$

$\Rightarrow \frac{4\sqrt{3}-5\sqrt{3}}{8\times 3}=-\frac{\sqrt{3}}{8\times 3}=-\frac{\sqrt{3}}{24}$.