Prove that $\lim_{x \to 0} \frac{1}{x}$ does not exist Here's what I'm trying to prove:
Prove that $\lim_{x \to 0} \frac{1}{x}$ does not exist.

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?
 A: Your proof is not OK.
You claim:


*

*That if the limit exists, then $f(x)$ is bounded for $|x|>\frac{1}{|L|+\epsilon}$.

*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.

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.

