# Prove that f approaches a limit near a if and only if a = 0

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1. Nov 3, 2015

### lordianed

1. The problem statement, all variables and given/known data
Let $f$ be defined by $f(x) = x$ if $x$ is rational, and $f(x) = -x$ if $x$ is irrational, prove that $\lim_{x\to a} f(x)$ exists if and only if $a=0$.

2. Relevant equations
Epsilon-delta definition of a limit: $\lim_{x\to a}f(x) = l$ means that for every $\epsilon > 0$ there exists a $\delta >0$ such that for all $x$, if $0<|x-a|<\delta$, then $|f(x)-l|<\epsilon$.

Its negation: There exists some $\epsilon >0$ such that for all $\delta >0$, there exists some $x$ for which $0<|x-a|<\delta$ but not $|f(x)-l|<\epsilon$.

3. The attempt at a solution

$(\Leftarrow)$
Suppose that $a = 0$, I guess guess that $\lim_{x\to 0}f(x) = 0$. Let $\epsilon > 0$ be arbitrary and choose $\delta = \epsilon$. For all $x$ that satisfy $0<|x|<\delta$, $x$ is either rational or irrational. If $x$ is rational, then $|f(x)| = |x| < \epsilon$; otherwise if $x$ is irrational, then $|f(x)| = |-x| = |x| <\epsilon$, then by definition $\lim_{x\to 0}f(x) = 0$.

$(\Rightarrow)$
Attempting a proof by contrapositive: Suppose that $a \neq 0$, and let $l$ be arbitrary, then it must be shown that for some $\epsilon >0$, for all $\delta >0$, there exists some $x$ for which $0<|x-a|<\delta$ but not $|f(x)-l|<\epsilon$.

This step is problematic for me,
I am unable to find such an $\epsilon$, I know that it most likely has to be related to $\delta$, but I am unable to systematically find it. Any help would be greatly appreciated :)

2. Nov 3, 2015

### Orodruin

Staff Emeritus
Hint: Any finite interval contains both rational and irrational numbers.

3. Nov 3, 2015

### Samy_A

There is a typo here, $\epsilon$ is most likely related to $a$.

4. Nov 3, 2015

### geoffrey159

Use the sequential definition of the limit, and density results

5. Nov 3, 2015

### lordianed

Thank you for the hints and correction Orodruin and Samy, I will mull over them; and thanks for your suggestion geoffrey, however the question is from Spivak's Calculus which does not cover sequences until much later, I'm sorry that I don't quite understand what you're telling me.

6. Nov 3, 2015

### PeroK

I would forget about $\epsilon - \delta$ first and look at what happens to the function near non-zero points. E.g. $x = \pm 1$, $x = \pm 10$, $x = \pm 0.1$. Try to see why the limit does not exist. Then, of course, you have to turn this understanding into a formal $\epsilon - \delta$ proof. But, if you don't understand why something is or is not true, then it's hard to produce an $\epsilon - \delta$ proof.

7. Nov 3, 2015

### lordianed

Thank you for the suggestion PeroK, I know that this may not sound very mathematically correct, but near any non-zero point a, the values of the function do not tend towards a singular value but instead we have the function get close to the negative and positive values of that point, where the points 'oscillate' between values near a and -a. In drawing this I got a graph that looked sorta like y=x and y=-x put together in the same graph, and it seemed to make sense to me that the limit at the non-zero points do not exist, however I am at a loss as to how to put this mathematically, perhaps I have to give it more thought.

8. Nov 3, 2015

### Samy_A

So, for $a<>0$, you will have points $x$ as close as you want to $a$ where $|f(a)-f(x)|$ will be "very close" to $|2a|$.
Now remember that you can choose $\epsilon>0$ based on the specific $a$ you are considering. What you have to do now is translate the "very close" from the previous sentence into a more formal argument.

9. Nov 3, 2015

### geoffrey159

Orodruin almost gave you the solution. The key point is that between any two real numbers you can find a rational number and an irrational number, and that these two numbers can be chosen as close as you wish.

EDIT: Remember that the last line of your $\epsilon - \delta$ proof should be $|\ell| < \epsilon$.

Last edited: Nov 3, 2015
10. Nov 3, 2015

### PeroK

Okay, so two things.

1) You understand the function, but you cannot (yet) map the behaviour of the function to the epsilon-delta regime. Nothing unusual there, everyone struggles with this to begin with.

2) If you look at your "oscillation" at each point, you may find the clue to the formal proof. Suppose a limit exists at $x = a$: call it $L$, say. Then, all function values for $x$ close to $a$ must be close to $L$. This means that all function values close to $a$ must be close to each other. Not necessarily as close as they are to the limit, but close to each other nevertheless. This property is called being a "Cauchy" sequence. You don't need to know this, but the concept is useful here. At $x = 10$, say, the function values do not get close to each other (some get close to $10$ and some get close to $-10$, so the limit cannot exist).

Here's a suggestion. Try to prove that for $x = 10$ that the limit is not $0$. This is just one specific case. The first step is to find an $\epsilon$. Can you guess a suitable $\epsilon$ for this case?

11. Nov 4, 2015

### lordianed

Here's what I have so far everyone, This is my approach to the idea that for $a=10$ the limit is not $0$. I take $\epsilon = 10$, for any $\delta > 0$, I know that I can find $x$ in the interval $0<x-10< \delta$, in which case $|x| > 10$.

For the general case, here is what I've been thinking. I have to prove that $\lim_{x \to a} \neq l$ for all $a \neq 0$ and all $l$. Upon observing I realise $\epsilon = |a|$ could be the required epsilon, and I've divided my proof into cases, where $a$ is positive or negative, and whether $l \geq 0$ or $l < 0$. Here's the idea:

Partial attempt at proof:
Suppose that $a > 0$.
Case 1, $l\geq 0$. Taking $\epsilon = |a|$, for any choice of $\delta$, we choose $x$ such that $0<x-a<\delta$; since in any real interval I can find rational and irrational $x$, I choose $x$ to be irrational, in which case $f(x) < -|a|$, and thus $|f(x) - l| > |l|+|a| \geq |a|$.
Case 2, $l < 0$. Take $\epsilon = |a|$, for all $\delta$ choose $x$ such that $0<x-a<\delta$ and is rational, then $f(x) > |a|$, and therefore $|f(x) - l| > |l|+|a| > |a|$.

Am I getting the right idea?

12. Nov 4, 2015

### PeroK

Yes, you're definitely getting the idea. I wouldn't worry if it seems quite pedestrian. I think it's a good idea to do the separate cases until you get the hang of things.

13. Nov 4, 2015

### lordianed

Thank you for the help everyone, I am very pleased to have received it! :) I have my proof written in full and will continue to do more questions.

14. Nov 4, 2015

### geoffrey159

Why don't you show once and for all that $\ell = 0$ ?
Since $f(a) \in \{ -a, a\}$, and since $a\in [a -\delta , a + \delta]$, then $|a| = |f(a)| = |f(a) -\ell | < \epsilon$.
And therefore $a = 0$