Prove that $\lim_{n \to \infty} n \int_0^{\infty} f(x) g(nx) dx = 0$ 
Let $f, g: (0,\infty) \rightarrow \mathbb{R}$ be non-negative continuous functions which are improper integrable on $(0, \infty)$ and such that
  $$
\lim_{x \to 0} f(x) = 0,\quad\text{and}\quad
\lim_{x \to \infty} x g(x) = 0.
$$
  Prove that
  $$
\lim_{n \to \infty} n \int_0^{\infty} f(x) g(nx) dx = 0.
$$

My attempt: I found a proof of
$\lim_{n \to \infty} n \int_1^{\infty} f(x) g(nx) dx = 0 $.
 A: After letting $t=nx$,
$$n \int_0^{\infty} f(x) g(nx) dx=
\int_0^{\sqrt{n}} f(t/n) g(t) dt+\int_{\sqrt{n}}^{n} f(t/n)g(t) dt
+\int_{n}^{\infty} f(t/n)g(t) dt.$$
Let $\epsilon>0$. Now we estimate the three terms on the right.
1) Since $\lim_{x \to \infty} x g(x) = 0$ there is $N_1>0$ such that for $n>N_1$, and $x\geq n$, we have $0\leq x g(x)\leq \epsilon$ and
$$\int_{n}^{\infty} f(t/n) g(t) dt=\int_{n}^{\infty} \frac{f(t/n)}{t} tg(t) dt\leq \frac{\epsilon n}{n}\int_{n}^{\infty}f(t/n) d(t/n)\leq \epsilon\int_0^{\infty}f(x) dx.$$
2) Since $g$ is integrable in $(0,+\infty)$ there is $N_2>0$ such that for $n>N_2$, we have $\int_{\sqrt{n}}^{+\infty} g(t) dt\leq \epsilon$ and
$$\int_{\sqrt{n}}^{n} f(t/n) g(t) dt\leq \max_{x\in [0,1]}f(x)\int_{\sqrt{n}}^{+\infty} g(t) dt\leq \epsilon \max_{x\in [0,1]}f(x)$$
3) Since $\lim_{x \to 0} f(x) = 0$ there is $N_3>0$ such that for $n>N_3$, and $0\leq x\leq 1/\sqrt{n}$, we have $0\leq f(x)\leq \epsilon$. Therefore
$$\int_0^{\sqrt{n}} f(t/n) g(t) dt\leq \epsilon \int_0^{\infty} g(t) dt.$$
A: Note by short $I=  \int_0^{\infty} f(x) g(nx) dx$. We have 
$$ n \int_0^{\infty} f(x) g(nx) dx=nI = \dfrac{x}{\frac{1}{\dfrac{nI}{x}}}$$ 
Tending to $\infty$ it takes the form $\dfrac{\infty}{\infty}$ so we can apply Hôspital's Rule and get the expression (formal) $$\frac{1}{D}\space \text {where } D=\frac{d}{dx}\left(\frac{1}{\dfrac{nI}{x}}\right)=\frac{-1}{D'}$$ where $$D'=\frac{-nI}{x^2}+\frac{n}{x}\frac{d}{dx}(I)\to 0$$ Thus the limit is equal to $0$ (because it has the form $\dfrac{-1}{\dfrac{1}{D'}}$ whit $D'\to 0$).
