Prove that $\lim_{x \to 0} \frac{1}{x}$ does not exist

Question

Here's what I'm trying to prove:

Prove that $\lim_{x \to 0} \frac{1}{x}$ does not exist.

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Proof Attempt:

Suppose that the limit does exist and denote it by $L$. Then, let $\epsilon > 0$. So:

$$\exists \delta > 0: 0 < |x| < \delta \implies |\frac{1}{x} - L| < \epsilon$$

In particular, we note that:

$$\frac{1}{|x|} < |L| + \epsilon$$

So, $f(x) = \frac{1}{x}$ is bounded when $|x| > \frac{1}{|L| + \epsilon}$. Now, let $\delta = \frac{1}{|L|+\epsilon}$. So, we have:

$$0 < |x| < \delta = \frac{1}{|L|+\epsilon}$$

$$\frac{1}{|x|} > |L| + \epsilon$$

$$|f(x)| > |L| + \epsilon$$

In other words, $f(x)$ is not bounded in this deleted neighbourhood of $x = 0$. Previously, we proved that if $f(x)$ does have a limit, then it is bounded on some neighbourhood of $x = 0$. So, our result above gives us a contradiction.

Hence, $L$ does not exist. This proves the desired assertion.

Can someone check if my proof above works or not? If it doesn't, how could I improve it?

Answer 1

Your proof is not OK.

You claim:

1. That if the limit exists, then $f(x)$ is bounded for $|x|>\frac{1}{|L|+\epsilon}$.
2. That $f$ is not bounded on a deleted neighborhood of $0$.

And then, from 1 and 2, conclude that $L$ does not exist. This is not OK. There is no clear logical connection between claims 1 and 2 and the conclusion. You should explain further why the conclusion follows.

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Also, in general, your attempt starts of saying "supposing that the limit exists". This suggests to the reader you are doing a proof by contradiction, however, there is later no point where you clearly say that you have now reached a contradiction. To make your proof clear to people who read it, it would be a good idea to explicitly explain where exactly the contradiction occurs.

In other words, proofs by contradiction usually have the form of:

Suppose $\neg A$ is true. ... ... ... Therefore, $B$ is true. However, we also know that $\neg B$ is true, which means we have reached a contradiction. We therefore conclude our initial assumption was false, and that $A$ must be true.