Question

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

Answer 1

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

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

= $\underset{x\to \sqrt{6}}{\lim }\left[\frac{\sqrt{5+2x}-\sqrt{3+2+2\sqrt{6}}}{\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)}\right]$

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

Rationalising the numerator:

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

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

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

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

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

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

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

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

= $\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{6}}$