Question

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

Answer 1

${lim}_{x\to \sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2x}-4}={lim}_{x\to \sqrt{2}}\frac{(x{)}^2-(\sqrt{2}{)}^2}{x^2+\sqrt{2x}-4}$

It is of the form $\frac{0}{0}$

$={lim}_{x\to \sqrt{2}}\frac{(x-\sqrt{2})(x+\sqrt{2})}{x^2+2\sqrt{2x}-\sqrt{2x}-4}$

$={lim}_{x\to \sqrt{2}}\frac{(x-\sqrt{2})(x+\sqrt{2})}{x(x=2\sqrt{2})-\sqrt{2}(x+2\sqrt{2})}$

$={lim}_{x\to \sqrt{2}}\frac{(x-\sqrt{2})(x+\sqrt{2})}{(x+2\sqrt{2})(x-\sqrt{2})}$

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

$=\frac{2}{3}$