Limits Of Exotic Functions

Calculus Level 3

$\begin{aligned} f(x) &= \begin{cases} 0 &\text{ if } x \text{ is irrational} \\ 1 &\text{ if } x \text{ is rational} \end{cases} \\ g(x) &= \begin{cases} 0 &\text{ if } x \text{ is irrational} \\ \frac1q &\text{ if } x =\frac{p}{q}, \text{ where } p \text{ and } q \text{ are coprime nonnegative integers} \end{cases} \end{aligned}$

Let $f(x)$ and $g(x)$ be two functions defined on $[0,1]$ by the formulas as described above.

For which $a \in (0,1)$ does the (deleted) $\lim\limits_{x\to a} f(x)$ exist?

For which $b \in (0,1)$ does the (deleted) $\lim\limits_{x\to b} g(x)$ exist?

All

a

; all

b

Irrational

a

; all

b

Irrational

a

; irrational

b

a

; all

b

a

; no

b

2 solutions

Patrick Corn
Jun 2, 2016

Every open interval contains some rational numbers and some irrational numbers, so $f(x)$ cannot have a limit anywhere; given any open interval, it will be $0$ at some points in the interval and $1$ at others.

But $\lim\limits_{x\to b} g(x) = 0$ for all $b \in (0,1).$ To see this, suppose we are given $\epsilon>0.$ Find a positive integer $N$ such that $1/N < \epsilon.$ There are finitely many rational numbers in $[0,1]$ with denominator $< N.$ If $b$ is not one of them, then we can find an open interval around $b$ that avoids all of them. If $b$ is one of them, we can find an open interval around $b$ that avoids all of them except for $b.$ In both cases, the value of $|g(x)-0|$ for all $x\ne b$ in the interval is less than $\epsilon.$ This is the requirement for the limit to equal $0.$

Drat. I interpreted the question as one of continuity rather than the weaker condition that the limit simply exists. So while $g(x)$ is continuous only for irrationals, the required limit does 'simply exist' for rationals as well, as per your proof. Nice question.

Brian Charlesworth - 5 years ago

Did the same here. And correct me if I am wrong, but isn't there a special name for both of these functions?

Anupam Nayak - 5 years ago

They're both considered to be pathological functions in general and Dirichlet functions in particular.

Brian Charlesworth - 5 years ago

@Brian Charlesworth – g is,sometimes, called Thomae's function...

vinod trivedi - 5 years ago

Same for me. Argh.

Tom Verhoeff - 4 years, 3 months ago

There's a slight error in the question statement (which doesn't affect the answer). The function $g(x)$ is not defined on $[0,1]$ by the given formula as claimed. Specifically $0$ is neither irrational nor expressible as a ratio of two positive integers, so $g(0$ is left undefined.

Erick Wong - 5 years ago

Good point. I will change "positive" to "nonnegative."

Patrick Corn - 5 years ago

I am unable to understand the g(x) part.

What I am thinking is if we consider b to be irrational then just right of it there are infinitely many rational as well as irrational points . So lim x tending to b+ can be anything either zero or some rational number . So limit should not exist.

Please clarify my doubt . @patrick corn

AYUSH JAIN - 5 years ago

But the value of $g(x)$ at those rational points gets arbitrarily small (with finitely many exceptions). See my proof above. Is there a particular part of the proof that you don't understand/agree with?

Patrick Corn - 5 years ago

yes , I am unable to understand the last line of your solution. also please explain be broadly how you are saying that they get very small. let me take an irrational number 0.73243421237654327543123....... now just right of it there could be a rational number say 0.732434210000000000000000001 this is a rational number whose 1/q is not infinitely small.

please point out what I am thinking wrong. Thanks sir.

AYUSH JAIN - 5 years ago

@Ayush Jain – The value of $g(x)$ at those rational numbers near your irrational number is nonzero, yes, but the value of $g(x)$ approaches $0$ as we get closer and closer to your irrational number, since the denominators of the rational numbers near $x$ have to increase as we get arbitrarily close.

Patrick Corn - 5 years ago

I strongly disagree on the definition of a limit. The value f(b) cannot be ignored in $\lim_{x\to b}f(x)$ . This is different from $\lim_{x\to b,x\neq b}f(x)$ .

Maxence Seymat - 2 years, 1 month ago

The limit of a function as $x\to b$ does not depend at all on the value of the function at $b.$ For more discussion, you can look here .

Patrick Corn - 2 years, 1 month ago

I also looked for details, and it seems that French mathematicians used mainly non-deleted limits, and that's the definition of a limit I worked with during my whole studies (I'm French by the way). Could you please specify this here for the answer depends on it?

Maxence Seymat - 2 years, 1 month ago

@Maxence Seymat – I checked French Wikipedia for limits and saw that it collaborated what you said.

Since it makes a significant difference to this problem, I've edited it in. Those who answered "No a, irrational b" have been marked correct, and I have removed that option.

Calvin Lin Staff - 2 years, 1 month ago

I'm not understanding why there are finitely many rational numbers in [0,1] with denominator < N. Why not infinitely many?

Atomsky Jahid - 5 years ago

For each $n<N$ make a list of all possible numerators: there is clearly a finite amount for each $n$ , so when you add up all the cases it is still finite. If you want to be specific, there are $\phi(n)$ possible numerators for each $n$ giving a total of $1+\sum_{n<N}^{}{\phi(n)}$ when we account for $0,1$ .

Dylan Pentland - 5 years ago

Thanks! It made sense. :)

Atomsky Jahid - 5 years ago

F(x) can have limit on $x\in (0,1)$ .

$\lim\limits_{x\to0.5} f(x) =1, \lim\limits_{x\to0.5^{+}} f(x) =1 , \lim\limits_{x\to0.5^{-}} f(x) =1$ ,

Because, $\lim\limits_{x\to0.5^{+}} x = 0.5000...1$ and $\lim\limits_{x\to0.5^{-}} x = 0.4999...$ which are rationals.

Akash Shukla - 5 years ago

Take the sequence $x_n = 0.5 + \frac{ \sqrt{2} } { 2^n }$ .
Does $x_n \rightarrow 0.5$ ?
What is $\lim f(x_n)$ ?

Note that when you want to prove a limit, it has to work for all sequences, and not just ones that you selectively choose.

Calvin Lin Staff - 5 years ago

I didn't understand why this has been asked? I'm not saying that there is limit of $f(x)$ for all $a$ . But what I have done,I think, I am right. About your question, $x_n→0.5$ is not possible. As $x_n ≠ 0.5^{-}$

Akash Shukla - 5 years ago

@Akash Shukla – Hm, are your familiar with limits of sequences ? Note that we do not need the limit to appear in the sequence. For example, the limit of $\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4} , \ldots$ is 0, even though 0 isnt' in the sequence.

I'm not sure why you said $\lim x_n = 0.5$ is not posisble because $x_n \neq 0.5$ .

Calvin Lin Staff - 5 years ago

@Calvin Lin – Yes, I am not familiar with limits of sequences. Sorry,for such a reply. I have written $0.5^{-}$ .

Akash Shukla - 5 years ago

@Akash Shukla – Are you trying to say "The limit of the sequence cannot be 0.5 because none of the terms are less than 0.5"?

If so, do you disagree that the limit of the reciprocals cannot be 0, because none of the terms are less than 0?
Along the same lines, do you think that the limit of 0, 0, 0, 0, 0, ... cannot be 0 because none of the terms are less than 0?

Calvin Lin Staff - 5 years ago

@Calvin Lin – Here lim is used, thats why I have written this. And I have said sorry for this.

Akash Shukla - 5 years ago

@Akash Shukla – Very nice problem. It seems to me that the key to Patrick's proof is the observation that "There are finitely many rational numbers in [0, 1] with denominator < N". Very nice!

John King - 5 years ago

It is a well known fact that there exist a rational number b/w 2 irrational numbers.

By using this I find both limits existing, for all x.

Aakash Khandelwal - 5 years ago

A Former Brilliant Member
Sep 5, 2019

The limit is a concept related to the neighbor of a point, not to the point itself. Even for infinity.

Take the example of the "f" function. Whenever "x" approaches the number "a" it finds either a rational number (f (x) = 1) or irrational (f (x) = 0). So the limit does not exist of any number "a".

Let's do the same analogy for the "g" function. But now we can find, each time that approaches the number "b", a rational number whose denominator "q" more and bigger. Then, the function "g" approaches zero for rational numbers and equal to zero for irrational numbers. So the limit exists for any number b

Limits Of Exotic Functions

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