2
$\begingroup$

I need to check whether the following functions are differentiable at 0, and if so if the derivative is continuous at 0.

  1. $f(x) = x^2\sin(1/x)$ if $x\not = 0$, $f(0) = 0$
  2. $f(x) = (1/x)\sin(x^2)$ if $x\not = 0$, $f(0) = 0$

In (1), $\lim_{x\to 0} {x^2\sin(1/x) - f(0)\over x^2 - 0} = \lim_{x\to 0}x\sin(1/x) = 0$. So it's differentiable. In (2), the limit as x approaches 0 is 1, so that is differentiable at 0 too.

I have two questions. How do I rigorously prove differentiability? Namely, how do I properly show the limits exist at 0?

Second, since both of these limits are constant at one point, 0, and are functions everywhere else, does that imply discontinuity? How do I check?

Thanks

$\endgroup$
  • $\begingroup$ Differentiability implies continuity. You need only show differentiability. $\endgroup$ – Wintermute Nov 6 '13 at 12:33
  • $\begingroup$ @mtiano I meant continuous at 0. That's what the problem states. Does that make a difference? $\endgroup$ – JohanLiebert Nov 6 '13 at 12:38
  • $\begingroup$ @JohanLiebert It just hit me why you're name is familiar. Monster is a great anime. $\endgroup$ – Git Gud Nov 6 '13 at 17:28
  • $\begingroup$ @GitGud Hehe, a true masterpiece :-) $\endgroup$ – JohanLiebert Nov 6 '13 at 17:37
1
$\begingroup$

To rigourously prove the first limit, note that $\sin$ is a bounded function and $x\to 0$ as $x$ approches $0$. Does this remind you of a proposition?

The second limit is $0$, not $1$. To see this note that $$0=\lim \limits_{x\to 0}(x)\cdot 1=\lim \limits_{x\to 0}(x)\cdot\lim \limits_{x\to 0}\left(\dfrac{\sin (x^2)}{x^2}\right)=\lim \limits_{x\to 0}\left(x\dfrac{\sin (x^2)}{x^2}\right)=\lim \limits_{x\to 0}\left(\dfrac{\sin (x^2)}{x}\right).$$

This proves that both functions are differentiable at $x=0$. This implies that they are continuous there. Points $x$ such that $x\neq 0$ offer no problem.

To check the continuity of the derivatives at $x=0$, simply find the lateral limits at $x=0$ of the derivatives and assess the situation.

$\endgroup$
  • $\begingroup$ What do you mean by lateral limits at $x=0$? I'm not familiar with this term. Could you show me what you mean? $\endgroup$ – JohanLiebert Nov 6 '13 at 17:38
  • $\begingroup$ @JohanLiebert Lateral limits are the limits taken from only one side of a point. For instance $\lim \limits_{x\to 0}\left(\dfrac 1 x\right)$ doesn't exist (finitely or infinitely) because from the left of $0$ it becomes $-\infty$ and from right it becomes $+\infty$, the notation is as follows for the limit from the right at $0$ is as follows: $\lim \limits_{x\to 0^\color{red}+}\left(\dfrac 1 x\right)=+\infty$. Having said this, I just checked and you don't need the concept of lateral limit. In my answer read 'simply find limits' and you're good to go. $\endgroup$ – Git Gud Nov 6 '13 at 17:44
  • $\begingroup$ @JohanLiebert Do you need further help with this? $\endgroup$ – Git Gud Nov 9 '13 at 12:07

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.