Question

Evaluate the limits: $\underset{x\to \sqrt{2}}{lim}\frac{x^2-2}{x^2+\sqrt{2}x-4}$

Answer 1

Given: $\underset{x\to \sqrt{2}}{lim}\frac{x^2-2}{x^2+\sqrt{2}x-4}becomes\frac{0}{0}form$

Using Factorization method

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

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

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

$[\because (a^2-b^2)=(a+b\left)\right(a-b\left)\right]$

$\Rightarrow \underset{x\to \sqrt{2}}{lim}\frac{(x+\sqrt{2})(x-\sqrt{2})}{x(x+2\sqrt{2})-\sqrt{2}(x+2\sqrt{2})}$

$\Rightarrow \underset{x\to \sqrt{2}}{lim}\frac{(x+\sqrt{2})(x-\sqrt{2})}{(x+2\sqrt{2})(x-\sqrt{2})}$

$\Rightarrow \underset{x\to \sqrt{2}}{lim}\frac{(x+\sqrt{2})}{(x+2\sqrt{2})}$

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

Therefore,

$\underset{x\to \sqrt{2}}{lim}\frac{x^2-2}{x^2+\sqrt{2}x-4}=\frac{2}{3}$