TITLE: If $\displaystyle \lim _{x\to +\infty}y(x)\in \mathbb R$, then $\lim _{x\to +\infty}y'(x)=0$
QUESTION [9 upvotes]: Not homework. I need this (or something similar) to solve 4. in this question.
Let $y:(a ,+\infty)\to \mathbb R$ be $C^1$.
Prove that $$\lim_{x\to +\infty}y(x)=\eta\text{ for some }\eta\in \mathbb R\implies\text{the following limit exists and } \lim_{x\to +\infty}y'(x)=0$$
Intuitively this is true because $\displaystyle \lim_{x\to +\infty}y(x)=\eta$ means that $y$ almost stops increasing or decreasing, so $\displaystyle \lim_{x\to +\infty}y'(x)=0$. But how to prove it?
I tried $$\lim_{x\to +\infty}y'(x)=0=\lim_{x\to +\infty} \lim_{h\to 0}\dfrac{y(x+h)-y(x)}h= \lim_{h\to 0}\lim_{x\to +\infty}\dfrac{y(x+h)-y(x)}h= \lim_{h\to 0}\dfrac{a-a}h=0,$$
but why I can change the order of the limits? If I can't even do that, how can I prove this?
After reading the threads Tyler provided, I now just need to prove that $\displaystyle \lim_{x\to +\infty}y'(x)$ exists? Please consider a suitable adaption to the  linked question.

REPLY [2 votes]: Here is another easy way to prove that $\lim f'(y)=0$ if this limit exists.
Suppose : $\displaystyle\lim_{y\to +\infty} f'(y)=\lambda\neq 0$. 
Without loss of generality, we can assume $\lambda>0$ (you can replace $f$ with $-f$ in the other case).
Then there is a $b>a$ such that $f'(x)>\dfrac\lambda 2$ for all $x\geqslant b$.
This leads you to : 
$$f(x)=f(0)+\int_0^x f' \geqslant f(0) + \frac\lambda 2 x,\ (\text{if}\ x\geqslant b)$$
But $f(0)+\dfrac\lambda 2 x$ tends to $+\infty$ as $x$ tends to $+\infty$ . 
This is contradictory, so $\displaystyle\lim_{y\to +\infty} f'(y)=0$