# 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 $[0,1]$?

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 $$[0,1]$$?

• What have you tried? As stated this looks like a question with no effort put in. May 12 at 17:26
• no f is diffrentiable at x=0 , see my solution May 12 at 17:47

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 $$x=0$$.

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

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

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

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

$$=0$$

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 $$[0,1]$$.