Define $f,g:\Bbb{R}\to \Bbb{R}$ by $$ f(x)=\begin{cases} x, & \text{if} \,\,\, x\in\Bbb{Q},\\ \sin(x), &\text{if}\,\,\, x\in\Bbb{R-Q}\end{cases}$$ and $$g(x)=\begin{cases} x \sin(x)\sin(1/x), & \text{if}\,\,\, x\ne0,\\ 0, & \text{if}\,\,\, x=0\end{cases}$$ at $x=0.$ how do I know if they are differentiable or not at $x=0$ as of $f$ it looks like related to Dirichlet's function and it is continuous everywhere so should $f$? And I find out \begin{align*}g'(0)&=\lim_{h\to 0}\frac{g(h)-g(0)}{h}\\ &=\lim_{h\to 0}\sin(h)\sin(1/h). \end{align*} Does this limit exist or not? I was thinking of $\sin(1/h)\le1$ so the limit is finite equal to $0$? thanks for any help.


Let's find out the derivative! At its heart, the derivative its defined as a limit. If we find out that both pieces of the function $f$ have the same derivative at $0$, then we can conclude that $f$ is derivable at $0$.

$$ f'(0)=\lim_{t\to 0} \frac{f(t)-f(0)}{t}= \begin{cases} \lim_{t\to 0} \frac{t}{t}=1, & \text{if $t$ is rational} \\ \lim_{t\to 0} \frac{sin(t)}{t}=\lim_{t\to 0} cos(t)=1, & \text{if $t$ is irrrational} \end{cases} $$

Thus, no matter whether we approximate ourselves to $0$ through the rational or irrational numbers, or a mix or both, the derivative of $f$ at $0$ is 1. This also implies that f is continuous at 0, as you suggested.

Now for $g$: $$ g'(0)=\lim_{t\to 0} \frac{g(t)-g(0)}{t}=\lim_{t\to 0} \frac{tsin(t)sin(1/t)}{t}=\lim_{t\to 0}sin(t)sin(1/t) $$ Which goes to 0 using your argument, which is indeed correct.

To see why it is correct, we have to use the sandwich rule. You can easily see that: $$ -sin(t)\le sin(t)sin(1/t)\le sin(t),\ \forall t\in \mathbb{R} $$ Which in the limit goes to: $$ 0\le\lim_{t\to 0}sin(t)sin(1/t)\le 0 $$ Thus, the limit is 0.

  • $\begingroup$ thanks for the answer@Jsevillamol (difficult to pronounce) don't mind $\endgroup$ – Onix Feb 4 '16 at 14:43

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.