I know the question is a weird title.

Define $f:\mathbb{R}\to\mathbb{R}$ by $$\begin{cases} 8x & \text{if $x$ is rational}\\ 2x^2+8 & \text{if $x$ is irrational} \end{cases}$$ Determine at which points of $\mathbb{R}$ the function $f$ has a limit and justify your conclusions.

So we see that it is possible that they may approach the same point at $x=2$. Since this is an accumulation point for both parts of the function we can inquire as to the existence of a limit here. We don't consider other points since there will clearly be no other accumulation points that might give the same limit.

I'm confused on how to show the limit formally exists though I am pretty sure it converges to 16 obviously. So I know I have something of the form $0<|x-2|<\delta$ and $|f(x)-16|<\epsilon$. Where I am stuck is talking about the function itself. I don't think showing this with sequences will work but I'm not sure.



The sequential definition of convergence will always work, as long as whatever limit you have actually converges to the point you are trying to prove it converges to. For this problem you might create two sequences $\{x_n\}, \{y_n\}$ such that $x_n \to 2$, $y_n \to 2$ and $x_n$ is always rational, $y_n$ is always irrational for all $n\in \mathbb{N}$. You can build on that if you think that approach would be easier. Other than that you could try to find a $\delta$ for each piece of $f$ in such a way that whenever $|x-2|<\delta$ then $|f(x)-8|<\varepsilon$ is true, regardless of whether $x$ is irrational of rational. If $x$ is rational, $\delta_r = \frac{\varepsilon}{8}$ (delta for rational $x$) seems like a good choice. If it's irrational, work the absolute values to coax something out. For example, $$\left|(2x^2+8)-16 \right| = \left|2x^2-8 \right| = 2\left|x^2-4 \right| = 2\left|x-2 \right| \left|x+2 \right|$$ In this way, you've made $|x-2|$ appear when $x$ is irrational. Can you bundle up the remaining $2|x+2|$ quantity inside a delta for irrational $x$? I'll call it $\delta_i$. When you're done, take $\text{min}\{\delta_r, \delta_i\} = \delta$ to guarantee $|f(x)-8|<\varepsilon$ will be satisfied for your $x$.

  • $\begingroup$ I'm all but certain you want $\min$ at the end. $\endgroup$ – Ian Oct 12 '15 at 2:03
  • $\begingroup$ @Ian I believe you are right! $\endgroup$ – graydad Oct 12 '15 at 2:04

Without using sequences, note that if $0<|x-2|<\delta \leq 1$ then $|2x^2+8-16|<c_1\delta$ and $|8x-16|<c_2\delta$. (I leave it to you to prove this and to identify $c_1,c_2$.) Let $c=\max \{ c_1,c_2 \}$; then if $|x-2|<\delta$ then (regardless of whether $x$ is rational or not) $|f(x)-16|<c\delta$.

There is also an elegant sequence argument, based on splitting into a rational subsequence and an irrational subsequence.


Let $0 < \varepsilon < 1$; if $x$ is rational, then $|f(x) - 16| < \varepsilon$ iff $16 - \varepsilon < 8x < 16+\varepsilon$; if $x$ is irrational, then $|f(x) - 16| < \varepsilon$ iff $16-\varepsilon < 2x^{2}+8 < 16 + \varepsilon$. In both cases, respectively we have $$ -\frac{\varepsilon}{8} < x-2 < \frac{\varepsilon}{8},\\ \sqrt{4-\frac{\varepsilon}{2}} - 2 < x-2 < \sqrt{4 + \frac{\varepsilon}{2}} - 2; $$ but these are implied by $$ |x-2| < \min \{ |\sqrt{4 - \frac{\varepsilon}{2}} - 2 |, \sqrt{4 + \frac{\varepsilon}{2} - 2}, \frac{\varepsilon}{8} \} $$

  • $\begingroup$ It is convenient to avoid such awkward tight bounds by simply assuming $\delta$ to be less than some fixed quantity for any $\varepsilon$. In my answer I used $1$. $\endgroup$ – Ian Oct 12 '15 at 2:30
  • $\begingroup$ @Ian: Thank you; I am aware of that. Sometimes I think, perhaps it is good to show beginners the line of thoughts, and then they can develop their own quicker ways to tackle such problems. I know some people with math degree who do not really understand how an epsilon-analysis goes, for they only memorize the typical "format" of an epsilon-argument... :) $\endgroup$ – Megadeth Oct 12 '15 at 2:35
  • $\begingroup$ Your first line is quite unclear. In particular, that iff in the first line is false. $\endgroup$ – user99914 Oct 12 '15 at 3:01
  • $\begingroup$ The clause before "and" is for rationals and that after "and" is for irrationals. @JohnMa Yeah I agree that it may not be clear to everyone. $\endgroup$ – Megadeth Oct 12 '15 at 3:04
  • 1
    $\begingroup$ @JohnMa: Yes, it is correct. I did not expect people would interpret this way, ha! Alright, since you pointed it out, I edited my answer to be readable by more people. $\endgroup$ – Megadeth Oct 12 '15 at 3:53

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