If $f(x)=x^2 \sin{\left ( \frac{1}{x^2} \right )}$ for $x \neq 0$ and $0$ for $x=0$, is $f$ differentiable on $[0,1]$?

Question

Consider the function

$$f(x)=\begin{cases} x^2 \sin{\left ( \frac{1}{x^2} \right )},\ \text{ x} \neq \text{0} \\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x} = \text{0} \end{cases}$$

Here is the graph of $f$

Since $$f'(x)=2x\sin{(1/x^2)}-\frac{2\cos{(1/x^2)}}{x}$$ we can see that $f'$ is not bounded below on $[0,1]$.

Is $f'$ defined at $0$?

Is $f$ is differentiable on $[0,1]$?

Answer 1

Firstly, for any non zero $x$, $f$ is diffrentiable. It is easy to prove. now, I am trying to prove for $x=0$.

$$f^{\prime}(0)=\lim_{x\to 0}\frac{x^2\sin(\frac{1}{x^2})}{x}=\lim_{x\to 0} x\sin(\frac{1}{x^2}).$$

Now, $RHL=\lim_{x\to 0^+} x\sin(\frac{1}{x^2})=0$ and $LHL=\lim_{x\to 0^-} x\sin(\frac{1}{x^2})=0$.

(Since $\sin(\frac{1}{x^2})$ is a bounded function so the limit is $0$ as $x\to 0$.)

Since both limits are equal, $f$ is differentiable at $x=0$.

Basically, $f$ is diffrentiable on $[0,1]$, but $f^{\prime}$ is not continious at $x=0$.

Answer 2

$$f(x)=\begin{cases} x^2 \sin{(1/x^2)}, \text{ x} \neq 0 \\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ x} = 0 \end{cases}$$

$$f'(0)=\lim\limits_{h \to 0} \frac{f(0+h)-f(0)}{h-0}$$

$$=\lim\limits_{h \to 0} \frac{h^2\sin{(1/h^2)}}{h}$$

$$=\lim\limits_{h \to 0} h \sin{(1/h^2)}$$

$$=0$$

Therefore, $f$ is differentiable at $0$.

For $x \neq 0$,

$$f'(x)=2x\sin{(1/x^2)}-\frac{2\cos{(1/x^2)}}{x}$$

$$\lim\limits_{x \to 0^+} f'(x) = -\infty$$ $$\lim\limits_{x \to 0^-} f'(x) = +\infty$$

Hence, $f'$ is not continuous at $0$.

$f'$ is defined at $0$, $f$ is differentiable on $[0,1]$.