Question

${lim}_{x\to 0}\frac{\sqrt{2}-\sqrt{1+cosx}}{x^2}$

Answer 1

${lim}_{x\to 0}\frac{\sqrt{2}-\sqrt{1+cosx}}{x^2}$

$={lim}_{x\to 0}\frac{\sqrt{2}-\sqrt{2co{s}^2\frac{x}{2}}}{x^2}$

$={lim}_{x\to 0}\frac{\sqrt{2}(1-cos\frac{x}{2})}{x^2}$

$={lim}_{x\to 0}\sqrt{2}\frac{\left(2si{n}^2\frac{x}{4}\right)}{x^2}$

$=2\sqrt{2}{lim}_{x\to 0}{\left(\frac{sin\frac{x}{4}}{\frac{x}{4}}\right)}^2\times \frac{1}{16}$

$[\because {lim}_{x\to 0}\frac{sin\theta }{\theta }=1]$

$=2\sqrt{2}\times \frac{1}{16}=\frac{1}{4\sqrt{2}}$