Question

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

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