Suppose $f(x)\geq 0$, and $\int_0^{+\infty} f^2(x)dx$ is convergent. Prove that $\lim\limits_{x \to \infty}\frac{\int_0^x e^t f(t) dt}{e^x}=0.$ Suppose $f(x)\geq 0$, and $\int_0^{+\infty} f^2(x)dx$ is convergent. Prove that $\lim\limits_{x \to \infty}\dfrac{\int_0^x e^t f(t) dt}{e^x}=0.$
Notice that we are not given the continuity of $f(x)$. Hence L' Hospital's rule can  not work here. If we consider apply AM-GM inequality, we obtain
$$e^{x}f(x)\leq \frac{f^2(x)+e^{2x}}{2},$$ where $\int_0^{+\infty} f^2(x)dx<+\infty$ but $\int_0^{\infty} e^{2t}=+\infty$, which gives nothing helpful.
How to solve it? Thanks. 

A proof from Kavi Rama Murthyk
Since $\displaystyle\int_0^{+\infty}f^2(x)dx$ is convergent，by Cauchy's convergence test, we have
$$\forall \varepsilon>0,\exists \xi>0,\forall x>\xi ~~~s.t.~~~ \int_{\xi}^{x} f^2(t)dt< 2\varepsilon^2.$$
Thus，as per Cauchy-Schwarz's inequality，we obtain
$$\int_{\xi}^x e^t f(t)dt \leq \left(\int_{\xi}^x f^2(t)dt \cdot \int_{\xi}^x e^{2t}dt\right)^{\frac{1}{2}}< \left(2\varepsilon^2 \int_{\xi}^x e^{2t}dt\right)^{\frac{1}{2}}=\varepsilon \left(e^{2x}-e^{2\xi}\right)^{\frac{1}{2}},$$
which implies
$$\frac{\int_{\xi}^x e^t f(t)dt}{e^x}\leq \varepsilon \left(1-e^{\frac{\xi}{x}}\right)^{\frac{1}{2}}\leq \varepsilon$$
holds for all $x>\xi$. Therefore， taking the limits of both sides as $x \to +\infty$, we have
$$\lim_{x \to +\infty}\frac{\int_{\xi}^x e^t f(t)dt}{e^x}\leq \varepsilon.\tag{1}$$
Meanwihle, notice that，for the fixed $\xi$,
$$\lim_{x \to +\infty}\frac{\int_0^\xi e^t f(t)dt}{e^x}=0.\tag{2}$$
$(1)$ plus $(2)$, we obtain
$$\lim_{x \to +\infty}\frac{\int_0^x e^t f(t)dt}{e^x}\leq\varepsilon,$$
by the arbitariness of $\varepsilon>0$，which implies
$$\lim_{x \to +\infty}\frac{\int_0^x e^t f(t)dt}{e^x}=0.$$
