Proving that if $\lim\limits_{x\to\infty} xf(x)=0$ and $\lim\limits_{x\to\infty}xf''(x)=0$ , then $\lim\limits_{x\to\infty}xf'(x)=0$. (using Taylor) Let $f ∈ C^2$  and $f:(0,∞)→R$ be a function such that $\underset{x \rightarrow \infty}{\lim}xf(x)=0$ and $\underset{x \rightarrow \infty}{\lim}xf''(x)=0$.
Prove that $\underset{x \rightarrow \infty}{\lim}xf'(x)=0$.
I somehow feel that this problem needs to be solved using Taylor expansion (taylor series) since it involves derivatives. However, I do not know how to show it.
 A: 
I use the following version of Lopital's rule: if $f(x),g(x)$ are differentiable on $(0,\infty)$, with $|g(x)|\to \infty $ as $x\to \infty$. Then $$\lim_{x\to \infty} \frac{f'(x)}{g'(x)} = L \implies \lim_{x\to \infty} \frac{f(x)}{g(x)} = L$$

Note that no assumption is imposed on $f(x)$. The proof is reminiscent to that of Stolz's theorem.

We use the lopital's rule on the function
$$\frac{e^x f'(x)-e^xf(x)}{e^x}$$
To obtain $$\lim_{x\to\infty} \frac{(e^x f'(x)-e^x f(x))'}{(e^x)'} = \lim_{x\to\infty} \frac{e^xf''(x)-e^xf(x)}{e^x} = \lim_{x\to\infty} (f''(x)-f(x))=0$$ Hence $$\lim_{x\to\infty}\frac{e^x f'(x)-e^xf(x)}{e^x} =0=\lim_{x\to\infty}(f'(x)-f(x))$$
This implies $\lim_{x\to\infty} f'(x) = 0$.  

Now use lopital's rule on 
$$\frac{xe^x f'(x)-xe^xf(x)}{e^x}$$
To obtain $$\begin{aligned}\lim_{x\to\infty} \frac{(xe^x f'(x)-xe^x f(x))'}{(e^x)'} &= \lim_{x\to\infty} \frac{e^xf'(x)+xe^xf''(x)-e^x f(x)-xe^x f(x)}{e^x} \\ &= \lim_{x\to\infty} (f'(x)+xf''(x)-f(x)-xf(x))=0 \end{aligned}$$ where we used result obtained above. Hence $$\lim_{x\to\infty}\frac{xe^x f'(x)-xe^xf(x)}{e^x} =0=\lim_{x\to\infty}(xf'(x)-xf(x))$$
This implies $\lim_{x\to\infty} xf'(x)=0$.
A: The theorem mentioned in this answer is the key here. Since $f, f''$ both tend to $0$ the derivative $f' $ also tends to $0$. Again note that if $g(x) =xf(x) $ then $g''(x) =xf''(x) +2f'(x) $ and thus $g, g''$ both tend to $0$. By the theorem in linked answer $g' =xf'+f$ tends to $0$ and we are done.
It is also clear from the above that the result holds under the weaker assumption that $xf''(x)$ is bounded as $x\to\infty$. 
