TITLE: Find the limit $\lim \limits_{n \to \infty}  \int_0^1 f(nx) \,dx $
QUESTION [8 upvotes]: Please try to help me with a question that I'm trying to solve.

$f(x)$ is continuous in the range of $[0, \infty)$ and $\lim \limits_{x \to \infty}x^2 f(x) = 1$.
Calculate $$\lim_{n \to \infty}  \int_0^1 f(nx) \,dx .$$

REPLY [9 votes]: Here's an argument based on l'Hôpital's rule:
$$
\begin{array}{rcl}
\lim_{u \to \infty} \int_0^1 f(ux) \, dx 
&\stackrel{\color{Brown}{ux = y}}{=}& \lim_{u \to \infty}\frac{\int_0^u f(y) \, dy}{u}
\\&\stackrel{\color{Brown}{\text{l'Hôpital}}}{=}& \lim_{u \to \infty} \frac{\frac{d}{du}\int_0^u f(y) \, dy}{\frac{d}{du}u} 
\\& \stackrel{\phantom{ux \to y}}{=}& \lim_{u \to \infty} f(u) 
\\&\stackrel{\phantom{ux \to y}}{=}& 0,
\end{array}
$$
since $u^2 f(u) \to 1$ as $u \to \infty$. 
In this proof, we use a general version of l'Hôpital's rule that is apparently not as well-known as it deserves to be. This requires only the denominator to approach $\infty$; it is not necessary that the numerator also go to infinity. This has been explained by Bill Dubuque in several posts in this site; e.g., see here and here. The linked posts contain a formal statement of the theorem.