Question

$\underset{x\to 0}{lim}\frac{1-\cos (1-\cos 2x)}{x^4}$

Answer 1

We have,

$\underset{x\to 0}{lim}\frac{1-\cos (1-\cos 2x)}{x^4}$

We know that

$\cos 2x=1-2{\sin }^{2}x$

Therefore,

$\underset{x\to 0}{lim}\frac{1-\cos (1-(1-2{\sin }^{2}x\left)\right)}{x^4}$

$\underset{x\to 0}{lim}\frac{1-\cos \left(2{\sin }^{2}x\right)}{x^4}$

$\underset{x\to 0}{lim}\frac{1-(1-2{\sin }^2({\sin }^{2}x\left)\right)}{x^4}$

$\underset{x\to 0}{lim}\frac{2{\sin }^2\left({\sin }^{2}x\right)}{x^4}$

$\underset{x\to 0}{lim}\frac{2{\sin }^2\left({\sin }^{2}x\right)}{{\sin }^{4}x}\times \frac{{\sin }^{4}x}{x^4}$

$2\underset{x\to 0}{lim}{\left(\frac{\sin \left({\sin }^{2}x\right)}{\left({\sin }^{2}x\right)}\right)}^2\times \frac{{\sin }^{4}x}{x^4}$

We know that

${lim}_{x\to 0}\frac{\sin x}{x}=1$

Therefore,

$\Rightarrow 2\times 1\times 1$

$\Rightarrow 2$

Hence, this is the answer.