Question

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

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

Answer 1

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

$\underset{x\to 0}{lim}=\frac{1-cos(1-cosx)}{x^4}$
$=\underset{x\to 0}{lim}\frac{2si{n}^2\left(\begin{array}{c}\frac{2si{n}^2\left(\begin{array}{c}\frac{x}{2}\end{array}\right)}{2}\end{array}\right)}{x^4}$
$=2\underset{x\to 0}{lim}\frac{si{n}^2\left(\begin{array}{c}si{n}^2\left(\begin{array}{c}\frac{x}{2}\end{array}\right)\end{array}\right){\left(\begin{array}{c}si{n}^2\left(\begin{array}{c}\frac{x}{2}\end{array}\right)\end{array}\right)}^2}{x^4{\left(\begin{array}{c}si{n}^2\left(\begin{array}{c}\frac{x}{2}\end{array}\right)\end{array}\right)}^2}$
$=2\underset{x\to 0}{lim}\frac{si{n}^4\left(\begin{array}{c}\frac{x}{2}\end{array}\right)}{{\left(\begin{array}{c}\frac{x}{2}\end{array}\right)}^4{2}^4}=\frac{1}{2^3}=\frac{1}{8}$