TITLE: Is my proof of $\lim_{x\to \infty}\frac 1x = 0$ correct?
QUESTION [6 upvotes]: I tried to prove $$\lim_{x\to \infty}\frac 1x = 0$$
I started as thus
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}$$
Applying L'Hospital's Rule
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}=\lim_{x\to \infty}\frac 1{2x}=\frac12\lim_{x\to \infty}\frac 1x$$
Thus,
$$\frac12\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac 1x$$
which therefore implies 
$$\lim_{x\to \infty}\frac 1x = 0$$
QED.

REPLY [5 votes]: What you have (very cleverly!) shown is that if the limit $\lim_{x\to\infty}{1\over x}$ exists, then, by L'Hopital, it can only equal $0$.  Simply Beautiful Art's answer establishes the same result for $\lim_{x\to\infty}x$.  The difference is, in your case the limit actually does exist, while in SBA's case it doesn't.  That was SBA's implicit message:  You haven't proven the limit is $0$, you've only proven a conditional statement; it remains to show that the limit exists.
One possible way to show that the limit exists without explicitly computing it would be to invoke (or prove) a theorem saying that a monotonically decreasing function that's bounded below necessarily has a limit as $x$ tends to infinity.
In essence, you've done the second step of a two-step process.  There are other MSE questions where assuming the limit exists allows you to compute it; when I have more time I'll try to provide some links.  This is the first time I can think of, though, where I've seen L'Hopital's rule used as part of the derivation.