Question

Evaluate the limit:
$\underset{x\to \sqrt{6}}{lim}\frac{\sqrt{5+2x}-(\sqrt{3}+\sqrt{2})}{x^2-6}$

Answer 1

We have,

$\underset{x\to \sqrt{6}}{lim}\frac{\sqrt{5+2x}-(\sqrt{3}+\sqrt{2})}{x^2-6}$

$=\underset{x\to \sqrt{6}}{lim}\frac{\sqrt{5+2x}-\sqrt{(\sqrt{3}+\sqrt{2}{)}^2}}{x^2-6}$

$=\underset{x\to \sqrt{6}}{lim}\frac{\sqrt{5+2x-}\sqrt{3+2+2\sqrt{3}\sqrt{2}}}{x^2-6}$

$=\underset{x\to \sqrt{6}}{lim}\frac{\sqrt{5+2x}-\sqrt{5+2\sqrt{6}}}{x^2-6}$ $\left(\frac{0}{0}\right)$form

On rationalising numerator, we get

$=\underset{x\to \sqrt{6}}{lim}\frac{(\sqrt{5+2x}-\sqrt{5+2\sqrt{6}})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}{(x^2-6)(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}$

$=\underset{x\to \sqrt{6}}{lim}\frac{({\left(\sqrt{5+2x}\right)}^2-{\left(\sqrt{5+2\sqrt{6}}\right)}^2)}{(x^2-6)(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}$

$=\underset{x\to \sqrt{6}}{lim}\frac{(5+2x-5-2\sqrt{6})}{(x^2-(\sqrt{6}{)}^2)(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}$

$=\underset{x\to \sqrt{6}}{lim}\frac{(2x-2\sqrt{6})}{(x^2-(\sqrt{6}{)}^2)(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}$

$=\underset{x\to \sqrt{6}}{lim}\frac{2(x-\sqrt{6})}{(x^2-(\sqrt{6}{)}^2)(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}$

$=\underset{x\to \sqrt{6}}{lim}\frac{2(x-\sqrt{6})}{(x-\sqrt{6})(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}$

$[(x-\sqrt{6})\ne 0]$

$=\underset{x\to \sqrt{6}}{lim}\frac{2}{(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}$

$=\frac{2}{(\sqrt{6}+\sqrt{6})(\sqrt{5+2\sqrt{6}}+\sqrt{5+2\sqrt{6}})}$

$=\frac{1}{\left(2\sqrt{6}\right)\left(\sqrt{5+2\sqrt{6}}\right)}$

$=\frac{1}{\left(2\sqrt{6}\right)\left(\sqrt{(\sqrt{3}+\sqrt{2}{)}^2}\right)}$

$=\frac{1}{\left(2\sqrt{6}\right)(\sqrt{3}+\sqrt{2})}$

$=\frac{(\sqrt{3}-\sqrt{2})}{\left(2\sqrt{6}\right)(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$

$=\frac{(\sqrt{3}-\sqrt{2})}{\left(2\sqrt{6}\right)({\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2)}$

$=\frac{(\sqrt{3}-\sqrt{2})}{\left(2\sqrt{6}\right)}$

Therefore,
$\underset{x\to \sqrt{6}}{lim}\frac{\sqrt{5+2x}-(\sqrt{3}+\sqrt{2})}{x^2-6}=\frac{(\sqrt{3}-\sqrt{2})}{2\sqrt{6}}$