Find the limit $\lim \limits_{n \to \infty}  \int_0^1 f(nx) \,dx $ 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 $$\lim_{n \to \infty}  \int_0^1 f(nx) \,dx .$$

 A: 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. 
A: First note that, since $f$ is continuous on $[0,\infty)$, it is integrable over any interval of finite length contained in $[0,\infty)$.
I assume you want to compute $$ \lim_{n\rightarrow \infty}\int_0^1 f(nx)\,dx.$$
The above limit is equal to (making the substitution $u=nx$):
$$
\lim_{n\rightarrow \infty}{1\over n}\int_0^n f(u)\,du.
$$
Since $\lim\limits_{x\rightarrow\infty}\bigl[ x^2 f(x)\bigr]=1$, for sufficiently large $x$, say $x\ge  A$:
$$
\tag{1}{1\over 2x^2}\le f(x)\le {2\over x^2}
$$
Now, for any number $C $
$$
\tag{2}\lim_{n\rightarrow \infty}{1\over n}\int_A^n {C\over x^2}\,dx=
\lim_{n\rightarrow \infty}{1\over n} \Bigl[{C\over A}-{C\over n}\,\Bigr]=0.
$$
From (1) and (2), it follows that $$\lim\limits_{n\rightarrow \infty}{1\over n}\int_A^n f(u)\,du=0.$$
Thus
$$\eqalign{
\lim_{n\rightarrow \infty}{1\over n}\int_0^n f(u)\,du
&=\lim_{n\rightarrow \infty}\Bigl[\,{1\over n}\int_0^A f(u)+{1\over n}\int_A^n f(u)\,du\Bigr]\cr
&=\lim_{n\rightarrow \infty}  \,{1\over n}\int_0^A f(u)\ +\ 
\lim_{n\rightarrow \infty}{1\over n}\int_A^n f(u)\,du \cr     
&=0.
}
$$
A: $\int_0^1 f(nx)dx = \frac{1}{n}\int_0^n f(x)dx$.  But we can easily show by the property of $f(x)$ that $\int_0^\infty f(x)dx = C$ exists, and that there is an $x_0$ such that $f(x)>0$ if $x>x_0$.  From that we determine that, when $n>x_0$.
$$\int_0^1 f(nx)dx = \frac{1}{n} \int_0^n f(x)dx < \frac{1}{n}\int_0^\infty f(x)dx = \frac C n$$. So for $\epsilon>0$ if $n>C/\epsilon$, $\int_0^1 f(nx)dx < \epsilon$.
Now you just need t show that the limit must be non-negative, and then you've shown the limit must be zero.
A: We have $\int_0^1f(nx)dx=\frac 1n\int_0^nf(t)dt$ after the substitution $t=nx$. We fix $\varepsilon>0$ and $A$ such that $|x^2f(x)-1|\leq \varepsilon$ if $x\geq A$. We have 
\begin{align*}
\left|\int_0^1f(nx)dx\right|&=\frac 1n\left|\int_0^1f(t)dt\right|+\frac 1n\left|\int_1^n\frac{t^2f(t)-1}{t^2}dt+\int_1^n\frac{dt}{t^2}\right|\\
&\leq \frac 1n\left|\int_0^1f(t)dt\right|+\frac 1n\left|\int_1^A\frac{t^2f(t)-1}{t^2}dt\right|+\frac 1n\int_A^n\frac{|t^2f(t)-1|}{t^2}dt+\frac 1n\int_1^{+\infty}\frac{dt}{t^2}\\
&\leq \frac 1n\left|\int_0^1f(t)dt\right|+\frac 1n\left|\int_1^A\frac{t^2f(t)-1}{t^2}dt\right|+\frac{\varepsilon}n\int_A^n\frac{dt}{t^2}+\frac 1n\\
&\leq \frac 1n\left(\left|\int_0^1f(t)dt\right|+\left|\int_1^A\frac{t^2f(t)-1}{t^2}dt\right|+\varepsilon+1\right).
\end{align*}
Now, one can find a $n_0$ such that if $n\geq n_0$ then $\left|\int_0^1f(nx)dx\right|\leq \varepsilon$, and the limit is $0$.
