Let $f$ be a differentiable Function on $(0,\infty)$ such that $\lim_{x\to \infty}\frac {f(x)}{x}$=0.
I have to show that $\liminf_{x\to \infty}|f'(x)|=0$.
I had thought to apply the mean value theorem for the function $g(x)= \frac {f(x)}{x}$ over the interval $(x,2x)$ so that $$ \frac{\frac {f(2x)}{2x}-\frac {f(x)}{x}}{x}=g'(c)=\frac{cf'(c)-f(c)}{c^2} $$ for $c\in (x,2x)$. Then $$ \frac {f(2x)-2f(x)}{2}=\frac{cf'(c)-f(c)}{c^2}=\frac {f'(c)}{c}-\frac {f(c)}{c^2} $$ How to proceed further I have no idea. Is my way is right? Or do I have to choose another way? Any help will be appreciated
Let $\epsilon>0$ be given. Then there exists $M_\epsilon>0$ such that $\left|\frac{f(x)}{x}\right|<\frac\epsilon3$ for all $x>M_\epsilon$. So for such $x$, the Mean Value Theorem tells us that there exists $\xi\in(x,2x)$ such that $$x|f'(\xi)|=|f(2x)-f(x)|\le|f(2x)|+|f(x)|<\frac\epsilon3\cdot 2 x+\frac\epsilon3\cdot x=\epsilon x.$$ Therefore $|f'(\xi)|<\epsilon$ for some $\xi \in(x,2x)$ whenever $x>M_\epsilon$. This makes $\liminf|f'(x)|\le \epsilon$. As this holds for all $\epsilon>0$, we get $\liminf |f'(x)|=0$.