Question

Solve:
$\underset{x\to 0}{lim}\frac{1-\cos x\sqrt{\cos 2x}}{x^2}$

Answer 1

We have,

$\begin{array}{c}{lim}_{x\to 0}\frac{1-\cos x\sqrt{\cos 2x}}{x^2}\\ ={lim}_{x\to 0}\frac{1-{\cos }^{2}x\cos 2x}{x^2\left(1+\cos x\sqrt{\cos 2x}\right)}\\ =\frac{1}{2}{lim}_{x\to 0}\frac{1-{\cos }^{2}x\cos 2x}{x^2}\\ =\frac{1}{2}{lim}_{x\to 0}\frac{1-{\cos }^{2}x\left(2{\cos }^{2}x-1\right)}{x^2}\\ =\frac{1}{2}{lim}_{x\to 0}\frac{1+{\cos }^{2}x-2{\cos }^{4}x}{x^2}\\ =\frac{1}{2}{lim}_{x\to 0}\frac{\left(-2{\cos }^{2}x-1\right)\left({\cos }^{2}x-1\right)}{x^2}\\ =\frac{1}{2}{lim}_{x\to 0}\frac{\left(2{\cos }^{2}x+1\right)\left(1-{\cos }^{2}x\right)}{x^2}\\ =\frac{1}{2}{lim}_{x\to 0}\frac{\left(2{\cos }^{2}x+1\right){\sin }^{2}x}{x^2}\\ =\frac{1}{2}\times 3\times 1=\frac{3}{2}\end{array}$

Hence, the answer is $\frac{3}{2}$.