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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.

Jennifer
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    Outline: use the definition of derivative to compute $f'(0)$, then compute $f'(x)$ for $x \neq 0$ and show $f'(x) \not\to 0$ as $x \to 0$. – user217285 Aug 10 '15 at 03:13
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    In other words using the definition $ f ' (0) = \lim_{x \to 0} \frac{f(x)-f(0) }{x-0}$ , we have to show that $\lim_{x \to 0} \frac{x^2 \sin\left( \frac 1 x \right) - 0 }{x-0}$ exists, and it does - it equals to zero. – john Jan 28 '21 at 06:31

1 Answers1

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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?)

David
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