3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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

2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

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.