Mathematics - Continuity - Differentiability Question with Solution | TestHub

${\lim }_{x\to 0}f(x)={\lim }_{x\to 0}\frac{x^2\log (\cos x)}{\log (1+x)}$ $={\lim }_{x\to 0}-\frac{x\cdot \log (\cos x)}{\frac{\log (1+x)}{x}}$ $={\lim }_{x\to 0}x\cdot \log (\cos x)=0\cdot \log 1=0$ $\because {\lim }_{x\to 0}f(x)=f(0)$ $\therefore f(x)$ is continuous at $x=0$ Now, ${\lim }_{x\to 0}\frac{f(x+h)-f(x)}{h}$ ${\lim }_{h\to 0}\frac{f(h)-f(0)}{h}={\lim }_{h\to 0}\frac{h^2\log \cos h-0}{\log (1+h)}$ ${\lim }_{h\to 0}\frac{h^2\log (\cos h)}{1}=0$ $\therefore f(x)$ is differentiable at $x=0$