Question

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

Answer 1

${lim}_{x\to \sqrt{2}}\left(\frac{x^2-2}{x^2+\sqrt{2}x-4}\right)={lim}_{x\to \sqrt{2}}\left(\frac{x^2-2}{(x^2-4)+\sqrt{2}x}\right)\times \left(\frac{(x^2-4)-\sqrt{2}x}{(x^2-4)-\sqrt{2}x}\right)={lim}_{x\to \sqrt{2}}\left(\frac{(x^2-2)\left[\right(x^2-4)-\sqrt{2}x]}{(x^2-4{)}^2-2x^2}\right)={lim}_{x\to \sqrt{2}}\left(\frac{(x^2-2)\left[\right(x^2-4)-\sqrt{2}x]}{x^4-8x^2+16-2x^2}\right)={lim}_{x\to \sqrt{2}}\left(\frac{(x^2-2)[x^2-4-\sqrt{2}x]}{x^2(x^2-8)-2(x^2-8)}\right)={lim}_{x\to \sqrt{2}}\left(\frac{(x^2-2)[x^2-4-\sqrt{2}x]}{(x^2-8)(x^2-2)}\right)=\left(\frac{2-4-2}{-6}\right)=\left(\frac{2}{3}\right)$