Question

The value of $f\left(0\right)$ so that the function $f\left(x\right)=\frac{1-\cos (1-\cos x)}{x^4}$ is continuous everywhere is

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

Answer 1

The correct option is A $\frac{1}{8}$

For the function to be continuous, the value of the function at $x=0$ must be equal to its limit at this point. So,
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{1-\cos (1-\cos x)}{x^4}=\underset{x\to 0}{lim}\frac{1-\cos \left(2{\sin }^2\frac{x}{2}\right)}{x^4}=\underset{x\to 0}{lim}\frac{2{\sin }^2\left({\sin }^2\frac{x}{2}\right)}{x^4}=\underset{x\to 0}{lim}\frac{2{\sin }^2\left({\sin }^2\frac{x}{2}\right)}{{\left({\sin }^2\frac{x}{2}\right)}^2}\cdot \frac{{\sin }^4\frac{x}{2}}{{\left(\frac{x}{2}\right)}^4}\cdot \frac{1}{16}=\frac{1}{8}$