# $f(x)=\begin{cases} 0 & \text{if$x$is irrat. } \\ \sin |x| &\text{if$x$is rat. }\end{cases}$, Show $\lim_{x \to x_0 }f(x) =0$ by Heine def.

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

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?

• 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 $|x| < \delta$ May 28, 2021 at 20:32

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.

https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences

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

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