I was given this function: $$ f(x)= \begin{cases} x^2, & x\in\Bbb Q\\ 9, & x\notin \Bbb Q \end{cases} $$

Is it non-differentiable at every $x \in \mathbb R$?

I think so and I wrote it as $f(x) = D(x)(x^2-9)+9$, where $D(x)$ is Dirichlet function and I know it is non-differentiable at every $x \in \mathbb R$, but in this example I can't say that for sure when $x = 3$ and $x = -3$.


2 Answers 2


You will have to check the points $x=3$ and $x=-3$ separately. In fact, by symmetry, it is sufficient to check one of these, say $x=3$. So we will have to see if $\lim_{x \to 3}\frac{f(x)-f(3)}{x-3}$ exists.

This limit does not exist. Indeed, take some sequence $(a_n)$ in $\mathbb R \setminus \mathbb Q$ that converges to $3$. Then $\lim_{n \to \infty} \frac{f(a_n)-f(3)}{a_n-3}= \lim_{n \to \infty} \frac{9-9}{a_n-3} = 0$. However, if we take some sequence $(q_n)$ in $\mathbb Q$ converging to $3$, then we have $\lim_{n \to \infty} \frac{f(q_n)-f(3)}{q_n-3} = \lim_{n \to \infty} \frac{q_n^2-9}{q_n-3}= 6$.

I will rephrase the argument not using sequences. Suppose that the limit exists. Since for any $\delta>0$, there exists a irrational number $a$ in the interval $]3-\delta,3+\delta[$ for which $\frac{f(a)-f(3)}{a-3}=0$, we have that the limit must be zero. But this interval also contains an rational number $q$, which we can pick such that $|q-3|<1$. Then $\frac{f(q)-f(3)}{q-3}=q+3$. Thus $\frac{f(q)-f(3)}{q-3}=q+3 > 2$ so the limit cannot be zero. This is a contradiction.

  • $\begingroup$ I didn't learn about sequences, so I try to contradicts this using the definition of limits. So for $x\in\Bbb Q$:$|x+3−L| \lt \epsilon$ and for $x\notin\Bbb Q$:$|−L| \lt \epsilon$, but I can't find an ϵ that will contradicts this. $\endgroup$
    – AviAsks
    Commented Jan 12, 2020 at 18:15
  • $\begingroup$ I rephrased my answer using the definition of the limit (then you see the argument is essentialy the same as the argument given by Julián). In any case, I think it is good to be able to work both with the definition of the limit and the characterisation using sequences. In this case, I think that the argument using sequences is a little neater. $\endgroup$
    – Krup'a
    Commented Jan 13, 2020 at 11:53

We say that $\lim_{x\rightarrow x_0} f(x)$ converges in $\mathbb{R}$ if $$\exists!L\in\mathbb{R}(\forall\epsilon>0\ \exists\delta>0\ \text{such that } \forall x \text{ satisfying } |x-x_0|<\delta \text{ there must happen that } |f(x)-L|<\epsilon).$$ The negation of this statement (to say that the limit doesn't exist) is as following $$\forall L\in\mathbb{R} \ \exists\epsilon>0\ \forall \delta>0 \ \exists x \text{ such that }|x-x_0|<\delta \text{ and } |f(x)-L|\geq\epsilon.$$

Therefore, to show that the limit $$\lim_{x\rightarrow \pm3} \frac{f(x)-9}{x\mp3}$$ doesn't exist (where $f(x)$ is your function), for all $L\in\mathbb{R}^*$ ($\mathbb{R}^*=\mathbb{R}-\{0\}$), take $\epsilon=|L|$. For all $\delta>0$, the interval $[\pm 3-\delta, \pm 3+\delta]$ always contains an irrational number, lets say $x_0$. In that case, $|\frac{f(x_0)-9}{x_0\mp3}-L|=|L|\geq \epsilon.$

Now, if $L=0$, take $\epsilon =2$, and $\forall\delta>0$ choose a rational number $x_1$ in the interval $[\pm 3-\delta, \pm 3+\delta]$ (if $x=3$, take $x_1>3$, and if $x=-3$, take $x_1<-3$). In this case, $|\frac{f(x_1)-9}{x_1\mp3}-L|=|x_1\pm 3|\geq\epsilon$.

Thereby, as the limit of interest doesn't exist for $x=\pm3$, your function is non-differentiable at every $x\in\mathbb{R}.$

  • $\begingroup$ You can't choose a specific number for x, you can only choose whether $x \in \Bbb Q$ or not. $\endgroup$
    – AviAsks
    Commented Jan 13, 2020 at 6:15
  • $\begingroup$ Yes, you can. In fact, you have to choose an adequate one in order to prove that the limit doesn't exist. $\endgroup$
    – Julian
    Commented Jan 13, 2020 at 6:24
  • $\begingroup$ But how do you show, for example, that $x=1$ does satisfy $|1-3| \lt \delta$? $\endgroup$
    – AviAsks
    Commented Jan 13, 2020 at 7:00
  • $\begingroup$ Your choosing must satisfy first the inequality $|x \mp 3|<\delta$. In general x=1 does not satisfy that inequality. If you prefer, take $x_1<-3$ when proving the limit does not exist for -3 and take $x_1>3$ in the other case. $\endgroup$
    – Julian
    Commented Jan 13, 2020 at 7:31
  • $\begingroup$ What I’m saying is that I don’t know whether that satisfies that inequality since I don’t know the value of $\delta$. How do you know x=1 doesn’t satisfy it? $\endgroup$
    – AviAsks
    Commented Jan 13, 2020 at 8:01

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