From $\int\limits_{1}^{+\infty} f(x)dx$ is convegent , we can say $\lim_{x\to +\infty}f(x)=0$?
I think it's true . Because some example for me it. $$\left(\int\limits_{1}^{+\infty}\dfrac{dx}{x^n};\int\limits_{1}^{+\infty}\dfrac{dx}{x^n.e^x};\cdots;\right)n>1$$ But I can't prove it.
Nope, it's not true! The problem is that the convergence of a limit doesn't depend on the height of a function, but the area underneath; if the function is large infinitely often, but with decreasing widths each time, the integral can still converge.
A nice simple example is
$$ f(x) = \begin{cases} 1 & x \in [n, n + 2^{-n}) \\ 0 & \text{otherwise} \end{cases} $$
where $n$ is restricted to the integers. It's easy to see that the integral is equal to the infinite sum
$$\sum_{n=1}^{+\infty} 2^{-n} $$
which does indeed converge.