Limits of a twice differentiable function. Let $f\in C^2(\mathbb{R})$ such that $\lim_{x\rightarrow\infty} xf(x)=0$ and $\lim_{x\rightarrow\infty}xf''(x)=0$. Prove that $\lim_{x\rightarrow\infty}xf'(x)=0$.

By the Taylor-Lagrange Formula for $b=x+1$ and $a=x$, we have
$$f(x+1)=f(x)+f^{\prime}(x)+\frac{1}{2}f^{\prime\prime}(c_x)$$
for some $c_x\in (x,x+1)$. We multiply by $x$:
$$xf(x+1)=xf(x)+xf^{\prime}(x)+\frac{1}{2}\frac{x}{c_x}c_xf^{\prime\prime}(c_x)$$
Now if $x\to +\infty$, $c_x\to +\infty$, and $c_xf^{\prime\prime}(c_x)\to 0$. We have $\displaystyle \frac{x}{x+1}\leq \frac{x}{c_x}\leq 1$, hence $\displaystyle \frac{x}{c_x}\to 1$. So the last term $\to 0$ as $x\to+\infty$. As $xf(x)$ and $xf(x+1)$ $\to 0$ also, we get that $xf^{\prime}(x)\to 0$. 
