1
$\begingroup$

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


$h$ 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 $h'(0)=0$.


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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I am sorry, I made a mistake in the bold text of my question, I need to show that h' is not continous in 0 (instead of what I first asked, to proof that h'(0) is continous in 0). Does the answer still work? If so, could you elaborate? $\endgroup$ – The Coding Wombat May 11 '18 at 15:57
  • 2
    $\begingroup$ @TheCodingWombat Yes note that h'(0)=0 but $\lim_{x\to 0} h'(x)$ doesn't exist. $\endgroup$ – user May 11 '18 at 15:58
  • $\begingroup$ So I guess I first have to proof that $h'(x) = 2x\sin(1/x) - \cos(1/x)$, if so, is there another way? $\endgroup$ – The Coding Wombat May 11 '18 at 16:32
  • $\begingroup$ It’s not difficult to calculate $f’(x)$ for $x\neq 0$ $\endgroup$ – user May 11 '18 at 17:48
  • 1
    $\begingroup$ Yes exactly you have proved that $f’(0)=0$ but lim $f’(x)$ at $0$ doesn’t exist. $\endgroup$ – user May 12 '18 at 10:20
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.