Is this a proper way of using L'Hôpital's rule? I was reading this solution from IMC-2017, and even though I understand the general idea, I have a question regarding the use of L'Hôpital's rule. 
Let $f:[0,+\infty) \to \mathbb{R}$ be a continuous function such that $\lim\limits_{x\to + \infty}{f(x)} = L$ (finite or infinite). 
We need to prove that $\lim\limits_{n\to \infty}{\int\limits_{0}^{1}{f(nx)dx}} = L$. 
One of the proposed solutions is to express $\int\limits_{0}^{1}{f(tx)dx}$ as $\frac{F(t)}{t}$ and then use L'Hospital's rule to show that the limit of integrals also equals $L$. 
How can we be sure that $\lim\limits_{t\to \infty}{F(t)} = \infty$ so that we can use L'Hospital? Does it somehow automatically follow from the fact that $\lim\limits_{x\to + \infty}{f(x)} = L$?
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 A: The hint is good: we have, with the substitution $tx=y$ (where $t>0$),
$$
\int_0^1 f(tx)\,dx=\frac{1}{t}\int_0^t f(y)\,dy=\frac{F(t)}{t}
$$
where
$$
F(t)=\int_0^t f(y)\,dy
$$
and so $F'(t)=f(t)$.
Now we want to compute
$$
\lim_{t\to\infty}\int_0^1 f(tx)\,dx=\lim_{t\to\infty}\frac{F(t)}{t}
$$
and l'Hôpital yields
$$
\lim_{t\to\infty}F'(t)=\lim_{t\to\infty}f(t)
$$
You need not worry about the limit of $F(t)$, because l'Hôpital applies when the denominator has infinite limit, independently on the limit of the numerator, which may even not exist (provided the other necessary assumption about the derivatives of the functions are satisfied, of course).
See the last-but-one paragraph in the proof given on Wikipedia.
A: $\lim_{x \to +\infty}f(x)=L \Rightarrow L-\varepsilon < f(x) < L+ \varepsilon$ if $x>x_0$. Then, (suppose $L>0$ and take $\varepsilon < \frac{L}{2}$) $(L-\varepsilon)x < F(x) < (L+ \varepsilon)x$ by monotonicity of integrals. Then, when $x$ goes to $+\infty$, using the lower bound, $F(x)$ has to go to infinity too.
The other cases are stated in a similar fashion. Hope it helps you get them managed. Cheers.
