Question

If the function
$f\left(x\right)=\{\begin{array}{c}\frac{2-x^2-2\cos x}{2x^4}\text{for}x\ne 0\\ k\text{for}x=0\end{array}$
is continuous at $x=0$ then the value of $k$ is

• A. $1$
• B. $0$
• C. $-\frac{1}{48}$
• D. $-\frac{1}{24}$

Answer 1

The correct option is D $-\frac{1}{24}$

Since $f\left(x\right)$ is continuous at $x=0$
$\therefore f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{2-x^2-2\cos x}{2x^4}=k\to \frac{0}{0}$
Apply L'hospital
$k=\underset{x\to 0}{lim}\frac{-2x+2\sin x}{8x^3}\to \frac{0}{0}$
$k=\underset{x\to 0}{lim}\frac{-2+2\cos x}{24x^2}\to \frac{0}{0}$
$k=\underset{x\to 0}{lim}\frac{-2\sin x}{48x}$
$k=-\frac{1}{24}$

Alternative solution
$\therefore f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)=k\underset{x\to 0}{lim}\frac{1-\frac{x^2}{2}-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-....)}{x^4}=k$
$k=-\frac{1}{4!}=-\frac{1}{24}$