The value of $\underset{x\to 0}{lim} \frac{1-\cos (1-\cos x))}{x^4}$, is

Solution:

We have, 

$\begin{array}{l}\underset{x\to 0}{lim} \frac{1-\cos (1-\cos x)}{x^4}=\underset{x\to 0}{lim} \frac{1-\cos \left(2{\sin }^{2}x/2\right)}{x^4}\\ =\underset{x\to 0}{lim} \frac{2{\sin }^2\left({\sin }^{2}x/2\right)}{x^4}=2\underset{x\to 0}{lim} {\left\{\frac{\sin \left({\sin }^{2}x/2\right)}{x^2}\right\}}^2\\ =2\underset{x\to 0}{lim} {\left\{\frac{\sin \left({\sin }^{2}x/2\right)}{{\sin }^{2}x/2}\times \frac{{\sin }^{2}x/2}{x^2/4}\times \frac{1}{4}\right\}}^2=2{\left(\frac{1}{4}\right)}^2=\frac{1}{8}\\ \begin{array}{rl}\text{ ALITER }& \underset{x\to 0}{lim} \frac{1-\cos (1-\cos x)}{x^4}\\ =& \underset{x\to 0}{lim} \left\{\frac{1-\cos (1-\cos x)}{(1-\cos x{)}^2}\right\}{\left(\frac{1-\cos x}{x^2}\right)}^2\\ =& \left(\frac{1}{2}\right)\times {\left(\frac{1}{2}\right)}^2=\frac{1}{8}\end{array}\end{array}$