Derivatives – Differentiable but Not Continuously Differentiable Functions

Question

Given $f: \mathbb{R} \rightarrow \mathbb{R}$ defined as
$$f(x)=\left\{\begin{array}{cc}x^2\sin\left(\frac{1}{x}\right)&,x\neq 0\\ 0&,x=0\end{array}\right\}.$$
I am trying to prove $f$ is differentiable at $x=0$ but not continuously differentiable there.

Answer 1

Here is the idea, I'll leave the detailed calculations up to you.

First, use normal differentiation rules to show that if $x\ne0$ then $$f'(x)=2x\sin\Bigl(\frac1x\Bigr)-\cos\Bigl(\frac1x\Bigr)\ .\tag{$*$}$$ Then use the definition of the derivative to find $f'(0)$. You should get $$f'(0)=0\ .$$ Then show that $f'(x)$ has no limit as $x\to0$, so $f'$ is not continuous at $0$. (Hint: the first term in $(*)$ tends to $0$; what happens to the second?)