Question

Evaluate the limit:
$\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7+2x}-(\sqrt{5}+\sqrt{2})}{x^2-10}$

Answer 1

We have,
$\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7+2x}-(\sqrt{5}+\sqrt{2})}{x^2-10}$

$=\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7+2x}-\sqrt{{(\sqrt{5}+\sqrt{2})}^2}}{x^2-10}$

$=\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7+2x}-\sqrt{5+2+2\sqrt{10}}}{x^2-10}$

$=\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7+2x}-\sqrt{7+2\sqrt{10}}}{x^2-10}$ $\left(\frac{0}{0}\right)$ form

On rationalising numerator, we get

$=\underset{x\to \sqrt{10}}{lim}\frac{(\sqrt{7+2x}-\sqrt{7+2\sqrt{10}})(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}{(x^2-10)(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}$

$=\underset{x\to \sqrt{10}}{lim}\frac{({\left(\sqrt{7+2x}\right)}^2-(\sqrt{7+2\sqrt{10}}{)}^2)}{(x^2-10)(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}$

$=\underset{x\to \sqrt{10}}{lim}\frac{(7+2x-7-2\sqrt{10})}{(x^2-10)(\sqrt{7+2x}+\sqrt{7+2}\sqrt{10})}$

$=\underset{x\to \sqrt{10}}{lim}\frac{(2x-2\sqrt{10})}{(x^2-(\sqrt{10}{)}^2)(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}$

$=\underset{x\to \sqrt{10}}{lim}\frac{(2x-2\sqrt{10})}{(x-\sqrt{10})(x+\sqrt{10})(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}$

$[(x-\sqrt{10})\ne 0]$

$=\underset{x\to \sqrt{10}}{lim}\frac{2}{(x+\sqrt{10})(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}})}$

$=\frac{2}{(\sqrt{10}+\sqrt{10})(\sqrt{7+2\sqrt{10}}+\sqrt{7+2\sqrt{10}})}$

$=\frac{2}{\left(2\sqrt{10}\right)\left(2\sqrt{7+2\sqrt{10}}\right)}$

$=\frac{2}{2\sqrt{10}\left(2\sqrt{{(\sqrt{5}+\sqrt{2})}^2}\right)}$

On rationalising denominator, we get

$=\frac{(\sqrt{5}-\sqrt{2})}{2\sqrt{10}(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}$

$=\frac{(\sqrt{5}-\sqrt{2})}{2\sqrt{10}(5-2)}$

Therefore,
$\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7+2x}-(\sqrt{5}+\sqrt{2})}{x^2-10}=\frac{(\sqrt{5}-\sqrt{2})}{6\sqrt{10}}$