Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{x\log \cos x}{\log (1+x^2)},& x\ne 0\\ 0,& x=0\end{array}$, then $f\left(x\right)$ is

• A. Continuous as well as differentiable at $x=0$
• B. Continuous but not differentiable at $x=0$
• C. Differentiable but not continuous at $x=0$
• D. Neither continuous nor differentiable at $x=0$

Answer 1

The correct option is A Continuous as well as differentiable at $x=0$

We have,
$L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}=\underset{h\to 0}{lim}\frac{-h\log \cos h}{-h\log (1+h^2)}$
$=\underset{h\to 0}{lim}\frac{\log \cos h}{\log (1+h^2)}\left(\frac{0}{0}\text{form}\right)$
$=\underset{h\to 0}{lim}\frac{-\tan h}{2h/(1+h^2)}=-1/2$
$R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{h\log \cos h}{h\log (1+h^2)}$
$=\underset{h\to 0}{lim}\frac{\log \cos h}{\log (1+h^2)}\left(\frac{0}{0}\text{form}\right)$
$=\underset{h\to 0}{lim}\frac{-\tan h}{2h/(1+h^2)}=\frac{-1}{2}$
Since $L{f}^{\prime }\left(0\right)=R{f}^{\prime }\left(0\right)$, therefore $f\left(x\right)$ is differentiable at $x=0$
Since differentiability $\Rightarrow$ continuity, therefore $f\left(x\right)$ is continuous at $x=0$.