Is my proof of $\lim_{x\to \infty}\frac 1x = 0$ correct? 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.
 A: 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.
A: Not an answer - essentially a comment and too long for a comment that I don't want lost in the flurry of existing comments.
Many students try L'Hopital unthinkingly when faced with the limit of an indeterminate form like $0/0$. Often the application is incorrect. Even when it works it's often not the easiest method, and it's rarely the most illuminating. You learn much more thinking about simple order of magnitude inequalities or the first few terms of Taylor series expansions.
There are many answers on this site that illustrate that. Here are some; other answerers should feel free to edit this answer to link to more. 
lHopitals $ \displaystyle \lim_{x\rightarrow \infty} \; (\ln x)^{3 x} $?
Finding the limit of a function with a trigonometric exponent
Computing $\lim_{x\to0} \frac 8 {x^8} \left[ 1 - \cos\frac{x^2} 2 - \cos\frac{x^2}4 + \cos\frac{x^2}2\cos\frac{x^2}4 \right]$ without using L'Hospital
Find the limit $\lim_{x\to0}\frac{\arcsin x -x}{x^2}$
A: I too tried the same thing:
$$\lim_{x\to\infty}x=\lim_{x\to\infty}\frac{x^2}x\stackrel{L'H}=2\lim_{x\to\infty}x$$
Thus,
$$\lim_{x\to\infty}x=2\lim_{x\to\infty}x$$
And as you have said,
$$\lim_{x\to\infty}x=0$$
QED (?)
A: 
The horizontal lines in the picture are $y = \pm \dfrac 12$. As you can see, after $P = 3$ on the $x$ axis, the values of $f(x)$ are contained on the interval $\left(-\dfrac 12, \dfrac 12 \right)$ on the $y$ axis. In informal terms, the rigorous definition of $\lim_{x \to \infty} \dfrac 1x = 0$ is simply the assertion that that you can do exactly what I did above for any horizontal lines $y = \pm \epsilon$, no matter what (positive) $\epsilon$ you pick. That is, for any positive number $\epsilon$, you can always find some point $P$ somewhere on the $x$ axis such that for every $x$ larger than $P$, its $f(x)$ value is on the interval $(-\epsilon, \epsilon)$.
A: This is incorrect, as you can only use L'Hospital's Rule when you know the limit of the derivative ratio exists.
A: Your expression In other words:
As x approaches infinity, then 1/x approaches 0 so its answer is 0
Try to think in that way...
Your method is wrong as you can only use L'Hospital's Rule when you know the limit of the derivative ratio exists.
