Question

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

Answer: (A)

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