Question

$\underset{x\to \sqrt{2}}{\lim }\frac{x^2-2}{x^2+\sqrt{2}x-4}$

Answer 1

$$\begin{gathered} \underset{x\to \sqrt{2}}{\lim }\left[\frac{x^2-2}{x^2+\sqrt{2}x-4}\right] \\ \mathrm{It}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{0}{0}. \\ \underset{x\to \sqrt{2}}{\lim }\left[\frac{x^2-{\left(\sqrt{2}\right)}^2}{x^2+2\sqrt{2}x-\sqrt{2}x-4}\right] \\ =\underset{x\to \sqrt{2}}{\lim }\left[\frac{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}{x\left(x+2\sqrt{2}\right)-\sqrt{2}\left(x+2\sqrt{2}\right)}\right] \\ =\underset{x\to \sqrt{2}}{\lim }\left[\frac{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}{\left(x-\sqrt{2}\right)\left(x+2\sqrt{2}\right)}\right] \\ =\frac{\sqrt{2}+\sqrt{2}}{\sqrt{2}+2\sqrt{2}} \\ =\frac{2\sqrt{2}}{3\sqrt{2}} \\ =\frac{2}{3} \end{gathered}$$