Let $f$ differentiable on $(a, \infty)$ such that $\lim_{x\to\infty}\frac{f(x)}{x}=0$. Show that $\lim\inf_{x\to\infty}|f'(x)|=0$ 
Let $f$ differentiable on $(a, \infty)$ such that $\lim_{x\to\infty}\frac{f(x)}{x}=0$. Show that $\lim\inf_{x\to\infty}|f'(x)|=0$. I would like to know if my proof holds, please and have a feedback.

As $\lim_{x\to\infty}\frac{f(x)}{x}=0$, we have by definition that: $\forall \epsilon>0 \ \exists x_0 \ \forall x\ge x_0$: $|\frac{f(x)}{x}|<\epsilon$.
Let $k \in (a,\infty)$. We have then $\forall \epsilon>0, \ \forall x\ge x_0$:
$|\inf|f'(x)||=\inf|f'(x)|\le|f'(x)|=\Big|\lim_{x\to k}\frac{f(x)-f(k)}{x-k}\Big|\le \lim_{x\to k}\Big|\frac{f(x)-f(k)}{x-k}\Big|< \lim_{x\to k}\epsilon=\epsilon$
So $\lim\inf_{x\to\infty}|f'(x)|=0$ as $k$ is arbitrary.
I'm not sure at all of this proof. The last inequaility seems to me false, but I didn't find a counter example. Another approach was to distinguish cases: $f$ converges or $f$ diverges (then apply L'Hôpital).
 A: Prove by contradiction. Suppose the contrary that $\liminf_{x\rightarrow\infty}|f'(x)|=l>0$.
Note that $\lim_{x\rightarrow\infty}\inf_{y\geq x}|f'(y)|=\liminf_{x\rightarrow\infty}|f'(x)|>l/2$,
so there exists $x_{0}$ such that $\inf_{y\geq x}|f'(y)|>l/2$ whenever
$x>x_{0}$. In particular, $|f'(y)|>l/2$ for all $y>x_{0}$.
Fix $x_{1}=x_{0}+1$. Let $x>x_{1}$ be arbitrary. By mean-value theorem, there
exists $\xi_{x}\in(x_{1},x)$ such that $|\frac{f(x)-f(x_{1})}{x-x_{1}}|=|f'(\xi_{x})|>l/2$.
On the other hand,
\begin{eqnarray*}
 &  & \frac{f(x)-f(x_{1})}{x-x_{1}}\\
 & = & \frac{\frac{f(x)}{x}-\frac{f(x_{1})}{x}}{1-\frac{x_{1}}{x}}.
\end{eqnarray*}
Note that as $x\rightarrow\infty$, $\frac{f(x_{1})}{x}\rightarrow0$,
$\frac{f(x)}{x}\rightarrow0$, and $\frac{x_{1}}{x}\rightarrow0$.
Therefore $\lim_{x\rightarrow\infty}\frac{f(x)-f(x_{1})}{x-x_{1}}=0$.
This contradicts to $|\frac{f(x)-f(x_{1})}{x-x_{1}}|>l/2$ whenever
$x>x_{1}$.
A: Ok, Your computations/intuition are good except that you are looking at the wrong limit by sending $x \to k$. You should send $x \to \infty$ because that is where you have some assumptions available. So, let me spell that out.
Given $\epsilon >0 $ I will show that for any $y$ there is a $c$ such that $y < c$ and $|f'(c)| < \epsilon$. I leave it to you to confirm that this implies $\liminf_{x \to \infty} |f'(x)| = 0. $
Given $\epsilon>0$ and $y>a$, we know that
$$
\lim_{x\to \infty} \frac{f(x)-f(y)}{x} = \lim_{x \to \infty} \frac{f(x)}{x} - \lim_{x \to \infty} \frac{f(y)}{x} = 0 \, .
$$
On the other hand,
$$
\lim_{x\to \infty} \frac{x}{x-y} = 1 \, .
$$
So, together they give
$$
\lim_{x \to \infty} \frac{f(x)-f(y)}{x-y} = \lim_{x \to \infty} \frac{f(x)-f(y)}{x} \cdot \frac{x}{x-y} = 0 \, .
$$
So, there exists an $x>y$ such that
$$
\left \vert \frac{f(x)-f(y)}{x-y} \right \vert < \epsilon \, .
$$
However, by mean value theorem we do have a $c \in (y,x)$ -- of course ultimately $x>y$ -- such that
$$
f'(c) = \frac{f(x)-f(y)}{x-y} \, .
$$
Therefore,
$$
|f'(c)| < \epsilon \, .
$$
So, to wrap up, for any $\epsilon >0$ and any $y>a$ I found a $c$ such that $ c>y$ and $|f'(c)| < \epsilon.$ Informally, this means that there will be arbitrarily small values for $|f'(x)|$ arbitrarily late at the tail.
