Let $f$ be a twice differentiable function on $ (0, \infty) $. Knowing that $$ \lim_{x\to\infty} x f(x) = 0 ,\;\;\;\; \lim_{x\to\infty} x f''(x) = 0.$$ Prove that $\lim_{x\to\infty} x f'(x) = 0$.
Current idea to finish this problem is to use Taylor's theorem and function $f(x+1)$ but my hour long attempts to do that have faced failure. So any help will be greatly appreciated.
You were on the right track. Let $x > 0$. Taylor's theorem gives you $$f(x+1) - f(x) = f'(x) + \frac{1}{2} f''(\xi)$$ where $\xi \in (x,x+1)$. Rearrange to get $$|f'(x)| \le |f(x+1)| + |f(x)| + \frac 12 |f''(\xi)|$$ which in turn leads to $$x |f'(x)| \le (x+1)|f(x+1)| + x|f(x)| + \frac{\xi}{2} |f''(\xi)|.$$ By hypothesis, given an $\epsilon > 0$ you can find $M > 0$ with the property that if $x > M$ then all three terms on the right are less than $\epsilon/3$.