Is my proof that this limit does not exist correct?

Question

Proposition:

Let $f$ be a function defined as $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise. The limit of $f(x)$ as $x$ approaches any number $a \neq 0$ does not exist.

My proof:

Suppose by contradiction that there is a limit $L$ for $f(x)$, as $x$ approaches $a \neq 0$. By definition of limit, there is a $\delta$-neighborhood of $a$ in which, as $x$ gets closer to $a$, in a $\epsilon$-neighborhood of $L$, $f(x)$ gets closer to $L$. Let's take a irrational number $x$ in such an neighborhood of $a$. That would imply $|L| < \epsilon$. We know that we can find a $y$ even closer to $a$ which implies that $f(y)$ is even closer to $L$ than $f(x)$ was. Since there are infinitely many numbers like that who happen to be rational, we have that $|y - L| < |L|$. But we also have infinitely many irrational numbers $z$, that are even closer to $a$ such that $f(z)$ is closer to $L$ than $f(y)$ was. So we must have $|L| < |y - L| < |L|$ which is absurd. We conclude then that there is no limit $L$. ■

Answer 1

I'm not sure I follow your proof entirely. you're using vague notions like "getting closer", and your logic is a bit wacky.

You want to disprove that there is a limit at $a$. The definition of limit at $a$ is

$\lim_{x\to a}f(x)$ exists $\iff$ there exists an $L$ such that for any $\epsilon>0$, there is a $\delta>0$ such that for any $x$ with $0<|x-a|<\delta$ we have $|f(x) - L|<\epsilon$

If you want to disprove this, then you have to prove the negation. We get:

$\lim_{x\to a}f(x)$ doesn't exist $\iff$ for any $L$, there exists an $\epsilon > 0$ such that for any $\delta > 0$ there is an $x$ with $0<|x-a|<\delta$ such that $|f(x) - L|\geq \epsilon$

Wow, that's a mouthful. However, the key part here is this: "there exists an $\epsilon >0$". You have to show that some $\epsilon$ exists. The easiest way of doing that is to pick one (cleverly), and then use that. Once I saw that you hadn't done that, I was almost certain that your proof wouldn't be correct.

For instance, a proof might go like this:

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Let $L$ be arbitrary. Set $\epsilon = \frac{|a|}3$. Let $\delta$ be arbitrary. Now pick a rational $x_1$ and irrational $x_2$ such that $0<|x_i-a|<\delta$ for $i = 1, 2$ and at the same time $|x_1|>\frac23|a|$. (For small $\delta$'s this last condition won't matter, as $|x_1-a|<\delta$ is a strictly stronger condidion, but one has to take into account that a devil's advocate might pick our arbitrary $\delta$ to be larger than $\frac13|a|$.) I will now show that $|f(x_1) - L|<\epsilon$ and $|f(x_2) - L|<\epsilon$ cannot both hold (and thus one of them must be $\!{}\geq\epsilon$).

Assume for contradiction that $|f(x_1) - L|<\epsilon$ and $|f(x_2) - L|<\epsilon$. We know that $x_2$ is irrational, which gives us $|L|<\frac{|a|}3$. We know that $x_1$ is rational, giving us $|x_1-L|<\frac{|a|}3$. Thus $$ |x_1 - L| + |L| < \frac23|a|\\ |x_1 - L + L| < \frac23|a|\\ |x_1| < \frac23|a| $$ which is a contradiction.

Answer 2

Your solution is not acceptable.

E.g. you use terminology like "closer" or "getting closer" without making clear what is meant by that. This makes it impossible to check your proof.

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Observe that $\mathbb Q^{\complement}$ and $\mathbb Q$ in $\mathbb R$ are both dense subsets of $\mathbb R$.

From that conclude that there are a sequences $(p_n)_n$ and $(q_n)_n$ both converging to $a$ with $p_n\in\mathbb Q^{\complement}$ and $q_n\in\mathbb Q$ for every $n$.

If a limit $L$ exists then $L=\lim_{n\to\infty}f(p_n)=a$ and $L=\lim_{n\to\infty}f(q_n)=0$ which is not possible if $a\neq0$.