limits - What is $lim_{x to infty} x^2f(x)$ where $int_0^infty f(x),mathrm{d}x = 1 $

I encountered a question in a test:

If $F(x)$ is a cumulative distribution function of a continuous non-negative random variable $X$, then find $\int_0^\infty (1 - F(x))\,\mathrm{d}x$

After a bit of pondering, I thought that the answer should depend upon the density of the random variable, so I checked the "none of these" option , but the correct answer was $\,E(X)$.So later I tried to work out the question properly.

If the density of random variable $X$ is $f(x)$ then it is necessary that $f(x) > 0$ and $\int_0^\infty f(x)\,\mathrm{d}x = 1$ Doing the integration by parts $$\left. x\Big(1-F(x)\Big)\right|_0^\infty - \int_0^\infty x\left(-\frac{\mathrm{d}}{\mathrm{d}x}F(x)\right)\,\mathrm{d}x$$ which reduces to $$ \lim_{x\to \infty} x\big(1\,- F(x)\big)\;+ \int_0^\infty xf(x) \, \mathrm{d}x $$

Now the $\int_0^\infty xf(x)\,\mathrm{d}x$ is clearly $E(X)$ but the limit is where my problem arises. Applying L'Hopital rule in the limit we have $$\lim_{x\to\infty}\frac{1-F(x)}{\frac{1}{x}} = \lim_{x\to\infty}\;x^2f(x)$$. Now is there any way to further reduce that limit to $0$ so that $E(X)$ is the correct answer or am I doing something wrong?