Question

Evaluate the limit:
$\underset{x\to \sqrt{2}}{lim}\frac{\sqrt{3+2x}-(\sqrt{2}+1)}{x^2-2}$

Answer 1

We have,
$\underset{x\to \sqrt{2}}{lim}\frac{\sqrt{3+2x}-(\sqrt{2}+1)}{x^2-2}$

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

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

$=\underset{x\to \sqrt{2}}{lim}\frac{\sqrt{3+2x}-\sqrt{3+2\sqrt{2}}}{(x-\sqrt{2})(x+\sqrt{2})}$ $\left(\frac{0}{0}\right)$ form

On rationalising numerator, we get

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

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

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

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

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

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

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

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


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

On rationalising denominator, we get

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

$=\frac{(\sqrt{2}-1)}{\left(2\sqrt{2}\right)({\left(\sqrt{2}\right)}^2-1)}$

Therefore,

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