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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 $\lim\limits_{x\to + \infty}{f(x)} = L$?

Link

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    $\begingroup$ LH can be applied when the denominator has infinite limit (it does not matter how the numerator behaves). $\endgroup$ – David Mitra Jan 25 at 11:29
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    $\begingroup$ L'Hospital's Rule is a powerful tool when used properly. It has become infamous because very few beginners try to learn its proper usage. The fact that it can applied for the form "$\text{anything} /\infty$" is not well known. $\endgroup$ – Paramanand Singh Jan 25 at 11:56
  • $\begingroup$ @ParamanandSingh, what's more, there are even some lecture notes and textbooks that (now I know that erroneously) claim that both the denominator and numerator should be $\infty$/$0$ $\endgroup$ – Don Draper Jan 25 at 12:01
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    $\begingroup$ Here is the rule of thumb : every serious student of calculus has to go through a lot of crappy textbooks before he/she really understands what's going on. $\endgroup$ – Paramanand Singh Jan 25 at 12:03
  • $\begingroup$ @DonDraper Do you mean that some sources claim it is wrong if only the denominator goes to $\infty$? Or do the "only" keep secret that there is a more general version? $\endgroup$ – Paul Frost Jan 25 at 12:20
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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.

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  • $\begingroup$ Thank you for your informative answer. It's interesting how I managed to never notice that the behavior of the numerator doesn't matter. $\endgroup$ – Don Draper Jan 25 at 16:19
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    $\begingroup$ @DonDraper Beware that this doesn't apply to the $0/0$ case: in order to apply l'Hôpital when the limit of the denominator is $0$, also the numerator must have limit $0$. $\endgroup$ – egreg Jan 25 at 16:21
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$\lim_{x \to +\infty}f(x)=L \Rightarrow L-\varepsilon < f(x) < L+ \varepsilon$ 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.

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