Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{6}$
• D. $\frac{1}{8}$

Answer 1

The correct option is D

$\frac{1}{8}$

Explanation for the correct option:

Compute the value of the given expression.

Given, $\underset{x\to 0}{\lim }\frac{\left[1-\cos (1-\cos x)\right]}{x^4}$

Simplify the given expression as follows:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\left[1-\cos (1-\cos x)\right]}{x^4}=\underset{x\to 0}{\lim }\frac{\left[1-\cos \left(2{\sin }^2\left({\displaystyle \frac{x}{2}}\right)\right)\right]}{x^4} \\ \Rightarrow =\underset{x\to 0}{\lim }\frac{2{\sin }^2\left({\sin }^2\left({\displaystyle \frac{x}{2}}\right)\right)}{x^4} \\ \Rightarrow =2\underset{x\to 0}{\lim }\frac{{\sin }^2\left({\sin }^2\left({\displaystyle \frac{x}{2}}\right)\right)}{x^4}·\frac{{\left[{\sin }^2\left({\displaystyle \frac{x}{2}}\right)\right]}^2}{{\left[{\sin }^2\left(\frac{x}{2}\right)\right]}^2}·\frac{2^4}{2^4}\left\{\text{Multiply numerator and demoninator by}{2}^4{\left[{\sin }^2\left(\frac{x}{2}\right)\right]}^2\right\} \\ \Rightarrow =2\underset{x\to 0}{\lim }\frac{{\sin }^2\left({\sin }^2\left({\displaystyle \frac{x}{2}}\right)\right)}{{\left[{\sin }^2\left(\frac{x}{2}\right)\right]}^2}·\frac{{\left[{\sin }^2\left({\displaystyle \frac{x}{2}}\right)\right]}^2}{x^4}·\frac{2^4}{2^4} \\ \Rightarrow =2\underset{x\to 0}{\lim }{\left[\frac{\sin \left({\sin }^2\left({\displaystyle \frac{x}{2}}\right)\right)}{{\sin }^2\left(\frac{x}{2}\right)}\right]}^2·\frac{{\left[{\sin }^2\left({\displaystyle \frac{x}{2}}\right)\right]}^2}{{\displaystyle \frac{x^4}{2^4}}}·\frac{1}{2^4} \\ \Rightarrow =2\underset{x\to 0}{\lim }{\left[\frac{\sin \left({\sin }^2\left({\displaystyle \frac{x}{2}}\right)\right)}{{\sin }^2\left(\frac{x}{2}\right)}\right]}^2·{\left[\frac{\sin \left({\displaystyle \frac{x}{2}}\right)}{{\displaystyle \frac{x}{2}}}\right]}^4·\frac{1}{2^4} \\ \Rightarrow =2·1^2·1^4·\frac{1}{2^4}\left\{\because \underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)=1\right\} \\ \Rightarrow =\frac{1}{8} \end{gathered}$$

Therefore, The value of $\underset{x\to 0}{\lim }\frac{\left[1-\cos (1-\cos x)\right]}{x^4}$ is $\frac{1}{8}$.

Hence, option (D) is the correct answer.