1
$\begingroup$

$$f(x) = \begin{cases} (x-2)^3 & \text{if $x$ is rational } \\ (2-x) & \text{if $x$ is irrational } \end{cases}$$

(i) Prove that if $c \ne 2$, then f does not have a limit at $x = c$.

(ii) Prove that $\lim_{x\to2} f (x)$ exists.

Hi all, i'm not very sure how to approach this question. For part i), i think i'm suppose to find a rational sequence $x_n$ and an irrational sequence $y_n$ such that $x_n \rightarrow c$ & $y_n\rightarrow c$. But $\lim_{n\to \infty}f(x_n) \ne \lim_{n\to \infty}f(y_n)$. Would letting $x_n = \frac{1}{n}$ and $y_n=\frac{1}{\sqrt{n}}$ suffice?

I'm clueless as to how to answer part ii). Would appreciate any hints or advice. Thanks in advance.

$\endgroup$
  • $\begingroup$ Your example for (i) almost works (sometimes $\sqrt{n}$ is rational), but it only applies for $c=0$. $\endgroup$ – vadim123 Apr 5 '15 at 3:24
  • $\begingroup$ The key insight is that if $\lim_{x\to c, x\in \mathbb R} g(x) = \ell$ for $g: \mathbb R \to \mathbb R$ continuous, then $\lim_{x\to c, x\in \mathbb Q} g(x) = \ell = \lim_{x\to c, x\in \mathbb Q^c} g(x)$. $\endgroup$ – William Stagner Apr 5 '15 at 3:28
2
$\begingroup$

For part $(i)$ you're on the right track, but note that we are concerned with $x_n, y_n$ converging to any arbitrary $c$. Note that your $x_n, y_n$ both converge to $0$ (also that $y_n$ is necessarily even a sequence of irrationals - consider $n = 4$). I'm guessing you are to come up with the sequences yourself, and I leave but a hint for you here:

  • If $c$ is rational consider adding to $c$ some rational sequence that converges to $0$ i.e. find an $a_n$ such that $\{a_n\} \subset \mathbb{Q}$ and $a_n \to 0$.
  • If $c$ is irrational consider first a sequence of decimal approximations for $x_n$ (that is the $n^{th}$ term of $x_n$ has $n$ decimal places expanded out) and for $y_n$ simply consider adding on the same sequence $a_n$ as above, will $c + a_n$ be irrational always?

Once you have done this then note that you will have $\lim_{n \to \infty} f(x_n) = (c - 2)^3$ and $\lim_{n \to \infty} = (2 - c)$. Now note $f$ is continuous at $c$ iff the limit is the same no matter how you approach it (meaning, no matter what sequence you use to approach $c$). If $c \neq 2$ what do you get above?

As for $(ii)$, we can use the sequential characterization of continuity. That is, $f : \mathbb{R} \to \mathbb{R}$ is continuous at $c$ iff for any $c_n \to c$ we have $$ \lim_{n \to \infty} f(c_n) = f(c) $$ Now theres three possible types of sequences when talking about rational and irrational sequences:

  • Completely rational sequence (all elements of the sequence are themselves rational)
  • Completely irrational sequence (all elements of the sequence are themselves irrational)
  • Mixed: Some elements are rational, some are irrational

Now when considering the case of $c = 2$ what is the limit of a completely rational sequence? What about a completely irrational sequence? As for this third category of sequences, the proof really depends on the level of rigor your professor/teacher wants. If you clue me in on how precise you want this to be I can help you out.

$\endgroup$
  • $\begingroup$ Hi @DanZimm, for part i), could i just use $x_n=c+\frac{1}{n}$ and $y_n=c+\frac{\sqrt{2}}{n}$? I've never really though about c being irrational or rational as i've only done questions where c is a whole number e.g 0,1,2 etc. Also for ii) could i use the $\varepsilon, \delta$ definition? $\endgroup$ – Helpisneeded Apr 5 '15 at 3:54
  • $\begingroup$ @Helpisneeded for $x_n$ defined as you put it, suppose $c = \pi$. Then is $\pi + \frac{1}{n}$ a rational number for every $n$? Similarly for your $y_n$ consider $c = - \frac{\sqrt{2}}{2}$, then it isn't true that $y_2$ is irrational (and thus not all of $y_n$ are irrational). As for the last part, since the sequence is made up of a combination of rationals and irrationals you can consider subsequences of rationals and irrationals, then show that any subsequence of $f(c_n) \to f(c)$ and thus $f(c_n) \to f(c)$. This may seem vague but working out the details yourself really helps you understand. $\endgroup$ – DanZimm Apr 5 '15 at 4:16
  • $\begingroup$ @Helpisneeded further, you can ditch the sequential formulation of continuity and just go from the definition - that is let $\epsilon > 0$ and show that there's a $\delta > 0$ so that $\lvert x - 2 \rvert < \delta \implies \lvert f(x) \rvert < \epsilon$. $\endgroup$ – DanZimm Apr 5 '15 at 4:17
  • $\begingroup$ Hi @DanZimm, sorry if this is taking up too much of your time, really appreciate the help. I see the problem with me trying to define $x_n$ and $y_n$. Do i then just let $x_n$ be a rational sequence where $x_n \rightarrow c$ and let $y_n$ be an irrational sequence where $y_n \rightarrow c$ and just say that $\lim_{n \to \infty} f(x_n) = (c - 2)^3$ and $\lim_{n \to \infty} f(y_n) = (2-c)$ and conclude that $c$ needs to be 2 for the limit to exist? $\endgroup$ – Helpisneeded Apr 5 '15 at 4:34
  • $\begingroup$ Also for the second part, could i just follow what @reluctant mathematician did below? $\endgroup$ – Helpisneeded Apr 5 '15 at 4:36
1
$\begingroup$

Since the rationals are dense in $\mathbb{R}$, for every $c\in\mathbb{R}$ there is a sequence $\{x_n\}$ of rationals which converges to $c$. Then the sequence $\{f(x_n)\}$ converges to $(c-2)^3$. Furthermore, since the irrationals are dense in $\mathbb{R}$, for every $c\in\mathbb{R}$ there is a sequence $\{y_n\}$ of irrationals which converges to $c$. Then the sequence $\{f(y_n)\}$ converges to $2-c$. Thus, the function can have a limit only where $(c-2)^3=2-c$, that is, at $c=2$.

To determine whether $f(x)$ has a limit at $x=2$, let $\epsilon>0$ and put $\delta=min\{\epsilon,\epsilon^{1/3}\}$. Let $x\in\mathbb{R}$ such that $|x-2|<\delta$. If x is rational, then $|f(x) - f(2)|=|(x-2)^3|<\delta^3$. If x is irrational, then $|f(x) - f(2)|=|2-x|<\delta$. In either case, $|f(x) - f(2)|<\epsilon$. $\ \ \ \ \Box$

$\endgroup$

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.