Prove that $\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}f'(x)$

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.

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

$$\lim_{x\to \infty}\frac{f(x)}{x}=\lim_{x\to \infty}f'(x)$$