Proving property of continuous random variables In proving the following property for a continuous random variable with $X\ge0$: $\int_0^\infty P(X>x)dx=E[X]$ where $E[X]$ is expectation of X
$\int_0^\infty P(X>x)dx=\int_0^\infty 1\times(1-P(X\le x))dx=x(1-F_X(x))|_0^\infty+\int_0^\infty xf_X(x)dx = x(1-F_X(x))|_0^\infty+E[X]$
Where $F_X(x)$ is the distribution function of X and $f_X(x)$ is the pdf of X.
Question 
How do I show that $x(1-F_X(x))|_0^\infty = 0$?
I have tried doing L'Hopital law i.e
$x(1-F_X(x))|_0^\infty=\lim_{n\to\infty}n(1-F_X(n))-0=\lim_{n\to\infty}\dfrac{1-F_X(n)}{\dfrac{1}{n}}=\lim_{n\to\infty}\dfrac{f_X(n)}{n^{-2}}$ by applying L'Hopital in the last step, however im still stuck in showing that this limit goes to $0$.
PS: (I have used the fact that by FTC derivative of distribution function is equal to pdf )
 A: *

*Way 1 : Since $X$ is a continuous r.v., it has a density function, i.e. there is $f_X$ s.t. $\mathbb P(X\in A)=\int_A f_X(x)dx$for all $A$ measurable. Therefore


\begin{align*}\int_0^\infty  \mathbb P(X>x)dx&=\int_0^\infty \int_x^\infty f_X(y)dydx\\
&\underset{Fubini}{=}\int_0^\infty f_Y(y)\int_0^ydxdy\\
&=\int_0^\infty yf_Y(y)dy\\
&=\mathbb E[X]
\end{align*}


*

*Way 2 : The formula hold in a more general case than just being continuous :
\begin{align*}
\int_0^\infty \mathbb P(X>x)\,\mathrm d x&=\int_0^\infty \int_{\{X>x\}}d\mathbb Pdx\\
&\underset{Fubini}{=}\int_\Omega  \int_0^{X(\omega )}dxd\mathbb P\\ &=\int_\Omega Xd\mathbb P\\
&=\mathbb E[X].
\end{align*}
A: Side note : your method and the standard solution given by Todd actually show that $x(1-F_{X}(x))\to 0$. 
This may also be seen directly by dominated convergence theorem :
$$x 1_{X\geq x} \le X 1_{X\geq x} \le X$$
and 
$$x 1_{X\geq x} \to 0 \text{ as } x \to \infty$$
imply
$$E[x 1_{X\geq x}]=x P(X\ge x) \to 0 $$
Nota bene : this beats by a little standard Markov inequality, that gives an effective bound, but not the $0$ limit, namely :
$$x P(X\ge x) \le E[X]$$
A: Given that $E(X)=\int_0^\infty x\,dF(x)$ exists, you have $$\lim_{t\to\infty}\int_t^\infty x\,dF(x)= 0$$ 
Again, 
\begin{align}
x(1-F(x))&=xP(X>x)
\\&=x\int_x^\infty dF(y)
\\&\le \int_x^\infty y\,dF(y)
\end{align}
So $\lim_{x\to\infty}x(1-F(x))\le 0$ and you already have $x(1-F(x))\ge 0$ for all $x\ge 0$.
Therefore a necessary condition for $E(X)$ to exist is that $$\lim_{x\to \infty} x(1-F(x))=0$$
A: This step
$$
\int_0^\infty 1\times(1-P(X\le x))dx=x(1-F_X(x))|_0^\infty+\int_0^\infty xf_X(x)dx 
$$
is not clear to me. I dont see what you did there, however if $f_X$ is the density of $X$ you have that
$$
\begin{align*}
\int_0^{\infty }\Pr[X>t]\,\mathrm d t&=\int_0^{\infty }\int_t^{\infty }f_X(s)\,\mathrm d s \,\mathrm d t\\
&=\int_0^{\infty }\int_0^{\infty }\mathbf{1}_{A}(s,t)f_X(s)\,\mathrm d s\,\mathrm d t\\
&\overset{(*)}{=}\int_0^{\infty }\int_0^{\infty }\mathbf{1}_{A}(s,t)f_X(s)\,\mathrm d t\,\mathrm d s\\
&=\int_0^{\infty }f_X(s)\left(\int_0^\infty \mathbf{1}_{A}(s,t)\,\mathrm d t\right)\,\mathrm d s\\
&=\int_0^{\infty }f_X(s)\left(\int_0^s \,\mathrm d t\right)\,\mathrm d s\\
&=\int_0^{\infty }f_X(s)s \,\mathrm d s\\&=E[X]
\end{align*}
$$
where $A:=\{(s,t)\in \Bbb R ^2:0\leqslant t<s\}$, and where we used Tonelli's theorem in $\rm(*)$.
