Question

$f\left(x\right)=\frac{1-\cos (1-\cos x)}{x^4}$ is continuous at $x=0$, then $f\left(0\right)=$

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

Answer 1

Answer: (D)

$\frac{1}{8}$

$f\left(x\right)=\frac{1-\cos (1-\cos x)}{x^4}$

It is given that the function $f$ is continuous at $x=0.$

Then, $\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$

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

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

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

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

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

$=\underset{x\to 0}{lim}\frac{1}{2^3}\frac{{\left(\sin {\left(\sin \frac{x}{2}\right)}^2\right)}^2}{{\left(\sin \frac{x}{2}\right)}^4}\underset{x\to 0}{lim}\frac{{\left(\sin \frac{x}{2}\right)}^4}{{\left(\frac{x}{2}\right)}^4}$

$=\frac{1}{8}\left(1\right)\left(1\right)$

$=\frac{1}{8}$

$\therefore$ $f\left(0\right)=\frac{1}{8}$