Show that the function $g(x) = x^2 \sin(\frac{1}{x}) ,(g(0) = 0)$ is everywhere differentiable and that $g′(0) = 0$ 
Show that the function $g(x) = x^2 \sin\left(\frac{1}{x}\right) ,(g(0) = 0)$ is everywhere differentiable and that $g′(0) = 0$.


We need to check differentiality only at $x=0$ since $x^2$ is differential everywhere and $\sin(\frac{1}{x})$ is differential everywhere except at $x=0$.
Now L.H.D.$= \lim_{h\to 0}\frac{h^2\sin(\frac{1}{h})-0}{h}=\lim_{h\to 0}h\sin(\frac{1}{h})=0$
Similarly, R.H.D.$=\lim_{h\to 0}\frac{(-h)^2\sin(\frac{1}{-h})-0}{-h}=\lim_{h\to 0}h\sin(\frac{1}{h})=0$ 
Since, L.H.D.=R.H.D., therefore, $g'(0)$ exists and equal to $0$. 
