Question

The value of $f\left(0\right)$ so that the function $f\left(x\right)=\frac{\cos x\left(\sin x\right)-\cos x}{x^4}$ is continuous at each point in its domain, is equal to

• A. $2$
• B. $-\frac{1}{4}$
• C. $\frac{1}{6}$
• D. $-\frac{1}{3}$

Answer 1

Answer: (C)

$\frac{1}{6}$

For$f$ to be continuous, we must have
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{\cos x\left(\sin x\right)-\cos x}{x^4}$

$=\underset{x\to 0}{lim}\frac{-2\sin x\frac{\sin x+x}{2}\sin \frac{\sin x-x}{2}}{x^4}$

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

$=-\frac{1}{2}\underset{u\to 0}{lim}\frac{\sin u}{u}\underset{v\to 0}{lim}\frac{\sin v}{v}\underset{x\to 0}{lim}(\frac{\sin x}{x}+1)\times \frac{lim\frac{-x^3}{3!}+\frac{x^5}{5!}+....}{x^3}$

$(u=\frac{\sin x+x}{2},v=\frac{\sin x-x}{2})$

$=-\frac{1}{2}\times 1\times 1\times 2\times \frac{-1}{3!}=\frac{1}{6}$