# Proving that limits exist and the derivative is continuous

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

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

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.
• What do you mean by lateral limits at $x=0$? I'm not familiar with this term. Could you show me what you mean? – JohanLiebert Nov 6 '13 at 17:38
• @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. – Git Gud Nov 6 '13 at 17:44