Differentiability of $h(x)=x^2\sin(1/x)$ for $x\in\mathbb{R} \backslash \{0\}$ and $h(0)=0$ and continuity of derivative

Question

Let $h(x)=x^2\sin(1/x)$ for $x\in\mathbb{R} \backslash \{0\}$ and $h(0)=0$.

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$h$ is differentiable in $0$ if $\displaystyle\lim_{k \to 0} \frac{h(0+k)-h(0)}{k}$ exists.

$$\lim_{k \to 0} \frac{h(0+k)-h(0)}{k}=\lim_{k \to 0} \frac{k^2\sin\left(\frac{1}{k}\right)}{k} = \lim_{k\to0} k\sin\left(\frac{1}{k}\right).$$

Using the Squeeze Theorem: $\left|\sin\left(\frac{1}{k}\right)\right|\leq1$, so $$\lim_{k\to0} -k = 0 \leq \lim_{k\to0} k\sin\left(\frac{1}{k}\right) \leq 0 = \lim_{k\to 0} k.$$ so $$\lim_{k\to0} k\sin\left(\frac{1}{k}\right)=0,$$ so $h$ is differentiable at $0$ and $h'(0)=0$.

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Is this proof of differentiability correct? I'm especially wary about the conclusion $h'(0)=0$, furthermore, how do I prove that $h'$ is not continuous at $0?$ (Can this be proven without proving what the derivative of $h$ is for all $x?$) I do not know where to start this continuity proof.

Answer 1

Yes the function is continuos in $x=0$ indeed

• $h(0)=0$ and as $x\to 0 \quad x^2\sin(1/x)\to 0$

and it is differentible at $x=0$ but since $h'(x)=2x\sin(\frac{1}{x})-\cos(\frac{1}{x})$ we have that $h'(x)$ is not continuos.

Answer 2

The proof of differentiability is correct. In fact you do not need to use Squeeze Theorem as $\lim_{x\rightarrow 0} f(x)g(x) = 0$ whenever $\lim_{x\rightarrow 0} f(x) =0$ and $g$ is bounded in a neighborhood of $0$. However, an easier approach to show that $h'$ is not continuous at $x =0$ is to consider the null sequence $\{\frac{1}{n\pi}\}$ and use the sequential criteria of continuity.