Question

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

• A. $\frac{1}{8}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. none of these

Answer 1

The correct option is A

$\frac{1}{8}$

${lim}_{x\to 0}f\left(x\right)=f\left(0\right)$ for continuity.
${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{1-cos(1-cosx)}{x^4}$
$cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}+.......\Rightarrow li{m}_{x\to 0}f\left(x\right)=li{m}_{x\to 0}\frac{1-[1-\frac{(1-cosx{)}^2}{2!}+\frac{(1-cosx{)}^4}{4!}....]}{x^4}\left(Replacingxby\right(1-cosx\left)\right){lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{(1-cosx{)}^2}{2!x^4}-\frac{(1-cosx{)}^4}{4!x^4}+....Also,{lim}_{x\to 0}\left(\frac{1-cosx}{x^2}\right)=\frac{1}{2}(\because cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}+.....){lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{1}{2}.{\left(\frac{(1-cosx)}{x^2}\right)}^2+\frac{1}{24}{lim}_{x\to 0}\frac{{(\frac{x^2}{2!}-\frac{x^4}{4!}.........)}^4}{x^4}+H.O.T.S{lim}_{x\to 0}f\left(x\right)=\frac{1}{2}\times {lim}_{x\to 0}{\left(\frac{1-cosx}{x^2}\right)}^2+0+0+.....{lim}_{x\to 0}f\left(x\right)=\frac{1}{2}\times {\left(\frac{1}{2}\right)}^2=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$