Mathematics - Continuity - Differentiability Question with Solution | TestHub

We have, $$\begin{array}{rl}{\mathrm{Lf}}^{\prime }(0)& =\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(0-\mathrm{h})-\mathrm{f}(0)}{-\mathrm{h}}=\underset{\mathrm{h}\to 0}{\lim }\frac{-\mathrm{h}\log \cosh }{-\mathrm{h}\log \left(1+{\mathrm{h}}^2\right)}\\ =& \underset{\mathrm{h}\to 0}{\lim }\frac{\log \cosh }{\log \left(1+{\mathrm{h}}^2\right)}\left(\frac{0}{0}\text{ form }\right)\end{array}$$ $\begin{array}{rl}& =\underset{\mathrm{h}\to 0}{\lim }\frac{-\tan \mathrm{h}}{2\mathrm{h}/\left(1+{\mathrm{h}}^2\right)}=-1/2\\ {\mathrm{Rf}}^{\prime }(0)& =\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{h}\log \cosh }{\mathrm{h}\log \left(1+{\mathrm{h}}^2\right)}\\ & =\underset{\mathrm{h}\to 0}{\lim }\frac{\log \cos \mathrm{h}}{\log \left(1+{\mathrm{h}}^2\right)}\left(\frac{0}{0}\text{ form }\right)\\ & =\underset{\mathrm{h}\to 0}{\lim }\frac{-\tan \mathrm{h}}{2\mathrm{h}/\left(1+{\mathrm{h}}^2\right)}=\frac{-1}{2}\end{array}$ Since ${\mathrm{Lf}}^{\prime }(0)={\mathrm{Rf}}^{\prime }(0),$ therefore $\mathrm{f}(\mathrm{x})$ is dif- ferentiable at $x=0$ Since differentiability $\Rightarrow$ continuity, there-fore $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$.