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On one of my textbooks has the following exercise:

Let $f$, $g$ and $h$ be the next functions:

$$f(x) = \left\{ \begin{array}{cc} 1& x \in \mathbb{Q}\\ 0 & x \in \mathbb{R}\setminus\mathbb{Q}\end{array}\right.\;\;\;\;\;\;\;\;\;\;g(x)=xf(x)\;\;\;\;\;\;\;\;\;\;h(x)=xg(x)$$

Prove that $g$ is continuous only on $x=0$ and $h$ is differentiable also only on $x=0$.

I know something about $f$: is a Dirichlet function and is not continuous (nor differentiable) everywhere.

I remember this matter from college and I think we proved this by showing that, for every $x_0\in\mathbb{R}$, the limit $\lim_{x\to x_0}f(x)$ is not well defined so we can't say that $\lim_{x\to x_0}f(x)=f(x_0)$ and, by this, we conclude that $f$ is not continuous on $x_0$.

Since is not continuous on $x_0$, $f$ is not differentiable on $x_0$ as well.

Here's my question: Doesn't the same argument show that $\lim_{x\to 0}g(x)$ is not well defined? Or that $\lim_{x\to 0}\frac{h(x)-h(0)}{x}$ is also not well defined? How can I state that $\lim_{x\to 0}g(x)=0$? Or $\lim_{x\to 0}\frac{h(x)-h(0)}{x}=0$?

A similar question has been asked here, however the answers don't provide any formal proof of this particular exercise (nor a leading clue to get it as far as I understand).

Thanks.

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Try the sandwich theorem and use $|f|\leq 1$, i.e. $$0\leq|g(x)|=|x||f(x)|\leq |x|\rightarrow 0 \mbox{ for }x\rightarrow 0.$$ Can you now do the same for the differential quotient of $h$?

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  • $\begingroup$ Yes, I can. I'm picture it now... Thanks. $\endgroup$ – Pspl Mar 5 at 15:49
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There is nothing about $\lim_{x\to x_0}f(x)$ which is not well-defined. What happens is that that limit doesn't exist, whatever the number $x_0$ is.

The same thing occurs with the function $g$ except if $x_0=0$. Then we have$$(\forall x\in\mathbb R):\bigl\lvert g(x)\bigr\rvert\leqslant\lvert x\rvert$$and therefore $\lim_{x\to0}g(x)=0=g(0)$.

A similar thing occurs with $h$.

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  • $\begingroup$ Actually, I meant that the limit doesn't exist (I shouldn't have say not well defined). Thanks for your answer, but I prefer to use the squeeze theorem as mentioned by @humanStampedist. But I understood yours as well. $\endgroup$ – Pspl Mar 5 at 15:55

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