Question says :

Let $f:\mathbb{R} \to \mathbb{R}$ defined as $f(x)=x^2 \sin(1/x^2)$ for $x \neq 0$ and $f(0) = 0$. Show that g differentiable $\forall x \in > \mathbb{R}$. Also show that the derivative $f'$ is unbounded on $[-1,1]$

For the first part :

For $x \neq 0$ $f(x)=x^2 \sin(1/x^2) \Rightarrow f'(x)=2x \sin(1/x^2) - \frac{2 \cos(1/x^2)}{x} $

For $x=0$ consider, $\lim_{x \to 0} \frac{f(x) - f(0)}{x-0}= \lim_{x \to 0} x\sin(1/x^2)$ $= 0 $ since $|x\sin{1/x^2} \leq |x| $ {sqeeze theorem}

now how do I show that this derivative function is unbounded on $[-1,1]$

  • 1
    $\begingroup$ The term $\frac{\cos(1/x^2) } x$ will cause problems near zero. $\endgroup$ – астон вілла олоф мэллбэрг Dec 2 '17 at 3:19
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    $\begingroup$ On the contrary, that term will provide a solution. $\endgroup$ – Henning Makholm Dec 2 '17 at 3:19
  • $\begingroup$ It is an interesting aspect of math that both the comments above are true although they seem contrary to each other. $\endgroup$ – Paramanand Singh Dec 2 '17 at 6:58

Let $x_{n}=\left(\dfrac{1}{2n\pi}\right)^{1/2}$ and put these $x_{n}$ into $2\cos(1/x^{2})/x$ to see what happens as $n\rightarrow\infty$.

  • $\begingroup$ And these $x_{n}$ make the term $\sin$ to be zero, glad. $\endgroup$ – user284331 Dec 2 '17 at 3:24

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