Prove that $\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}f'(x)$ I have come up with a theorem. (It probably doesn't qualify as a theorem, but I don't know what else to call it) This is what it states:
If 
$$\lim_{x\to\infty}f'(x)$$ exists (is finite), then $$\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}f'(x)$$
I have yet to find a counterexample, and it just seems to make sense to me, but I can't actually prove it.
This "theorem" is crucial to an answer to a question that I was asked to give, but my answer won't be a very good one if I can't prove why it works.

What I've tried:
I have been able to prove it if $f(x)$ has a linear asymptote (I think my terminology is correct).
If:
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}mx+b$$
Then:
$$\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}\frac{mx+b}x=m$$
But that doesn't work if there is no asymptote. Take $f(x)=\ln(x)$ for example. the $b$ in $mx+b$ would be infinite.
 A: A rigorous way to show this is by invoking $\varepsilon$-$M$ arguments. By 
condition $c \equiv \lim_{x \to \infty} f'(x)$ exists and finite, which implies that for any given $\varepsilon > 0$, there exists $M > 0$ such that for all $x \geq M$, 
$$|f'(x) - c| < \varepsilon. \tag{1}$$
On the other hand, by the mean value theorem, for any $x > M$, we have
$$f(x) = f(M) + f'(\xi)(x - M)$$
for some $\xi \in (M, x)$. Therefore, for all $x > M$, it follows that
\begin{align*}
& \left|\frac{f(x)}{x} - c\right| \\
= & \left|\frac{f(M) + f'(\xi)(x - M)}{x} - c\right| \\
= & \left|f'(\xi) - c + \frac{f(M) - f'(\xi)M}{x}\right| \\
\leq & |f'(\xi) - c| + \frac{|f(M) - f'(\xi)M|}{x} \\
< & \varepsilon + \varepsilon = 2\varepsilon.
\end{align*}
In the last step we may increase $x$ whenever necessary so that the latter term is bounded by $\varepsilon$, it is possible to do so since the numerator is bounded by a fixed number. This completes the proof.
A: 
L'Hopital's Rule does not require the numerator to approach infinity.  In fact, the limit $\lim_{x\to \infty}f(x)$ need not even exist.  See the note HERE that references the case of interest.

Thus, if $f'(x)$ exists for some interval $(x_0,\infty)$ and $\lim_{x\to \infty}\frac{f'(x)}{1}$ exists, then we have
$$\lim_{x\to \infty}\frac{f(x)}{x}=\lim_{x\to \infty}f'(x)$$
A: The following is not a rigorous answer (good ones were posted already), but rather a note on the intuition behind the result, somewhat related to the last part that you edited into the question.
When the limit exists $\lim_{x\to\infty}f'(x) = L$, intuitively $f(x)$ gets closer and closer to a linear function of slope $L$ as $x\to \infty$ (which can be a horizontal asymptote if $L = 0$ or an oblique one otherwise).
Now, consider a point $P \equiv (x,y)$ on the curve $y = f(x)$. The line through the origin $O \equiv (0,0)$ and point $P$ has the slope $y / x = f(x) / x$. For large enough $x$, the line $OP$ will intersect the asymptote of $f(x)$ in a vicinity of $P$, which becomes smaller and smaller as $x$ grows towards $\infty$ since $f$ "approaches" its asymptote. At the limit $x\to \infty$ the intersection point itself tends to infinity, which is to say that $OP$ tends to a line parallel to the asysmptote i.e. $\lim_{x\to \infty}f(x) / x = L = \lim_{x\to \infty}f'(x)$.
