Question

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

Answer 1

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

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

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

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

Rationalising the numerator:

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

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

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

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

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

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

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

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

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

= $\frac{\sqrt{2}-1}{2\sqrt{2}}$