Question

Evaluate

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

• A. $4$
• B. $2$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $2$

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,

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

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

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

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

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

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

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.