$ f(x)=\begin{cases} 0 & \text{if $x$ is irrational} \\ \sin |x| &\text{if $x$ is rational}\end{cases} $

Let $ x_0 \in \{ \pi n : n \in \Bbb Z \} $. Show that $ \lim_{x \to x_0 }f(x) $ exists or not, if it exists, find it.

Attempt: Let $ \epsilon > 0$ be arbitrary. Since we know that $ \sin |x| $ is continuous then there exists $ \delta >0 $. Let $ x \in \Bbb R $ be arbitrary. Suppose $ | x- x_0 | < \delta $. Then we know $ | \sin|x| | <\epsilon $ ( since $ \lim_{x \to 0 } \sin |x| = 0 $ ). Now,
If x is irrational, $ f(x) = 0 $ and so $ |f(x)| = 0 < \epsilon $.
If x is rational,$f(x) = \sin|x| $ and so $ |f(x)| = |\sin |x| | < \epsilon $.
Since $ \epsilon>0, x \in \Bbb R $ were arbitrary, we showed that $ \lim_{x \to x_0 }f(x) = 0 $. $ \square $

Question: Initially I tried to solve the problem using Heine's definition of limit but I found it to be very difficult. How would one prove the limit above using Heine's definition?

  • $\begingroup$ I may have mistaken In my proof attempt, because I am looking at an $ x $ s.t. $ | x - x_0 | < \delta $. I used the fact that $ \lim_{x \to 0 } \sin |x| = 0 $ but that is for $ |x| < \delta $ $\endgroup$ May 28, 2021 at 20:32

2 Answers 2


You can consider three cases. Let $x_1, x_2, \dots$ converge to $x_0$. We need to show that $f(x_1),f(x_2),\dots$ converges to zero (i.e. that it is a null sequence).

First suppose that the sequence of $x_j$ is eventually all irrational. Then, the sequence $f(x_j)$ is eventually all zero.

Second suppose that the sequence of $x_j$ is eventually all rational. Then, the sequence $f(x_j)$ is eventually $f(x_j) = \sin|x_j|$. So this goes to zero by the continuity of $x \mapsto \sin|x|$.

Lastly it may be that we have rationals and also irrationals appearing as $x_j$ infinitely often. Then we have finite streaks of zeros in the sequence $f(x_j)$. It is known in the basic theory of sequences that any sub-sequence of a convergent-to-$x_0$ sequence is also convergent-to-$x_0$. So those $x_j$ that are rational also converge to $x_0$. The corresponding sub-sequence of $f(x_j)$ thus is a null sequence, again by the continuity of $\sin|x|$. Lastly we apply (to the $f(x_j)$) the theorem that if any given sequence be partitioned into two sub-sequences, both of which are null, then the full sequence is also null.

That satisfies Heine's version. Your proof via Cauchy's definition looks okay; there are maybe some improvements possible to the phrasing.



First note that $|f(x+n\pi)| = |f(x)|$ so we can assume without loss of generality that $x_{0}=0$

Let $\varepsilon>0$. We need to find $\delta>0$ such that $|x|<\delta \implies |f(x)|<\varepsilon$

Well $|f(x)|\leq|\sin(|x|)|\leq|x|$

So taking $ \delta = \varepsilon $ gives the result

  • $\begingroup$ This seems correct to me. $\endgroup$ May 28, 2021 at 20:39
  • 1
    $\begingroup$ -1 The OP asks specifically how to prove it using Heine’s definition of limits, i.e. with arbitrary sequences instead of $\epsilon$-$\delta$ $\endgroup$
    – Milten
    May 29, 2021 at 21:12

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