Slope at $-\infty$ Given a twice differentiable function  $f(x)$ on $\mathbb R$ with the following properties:


*

*$f$ is an increasing function in $\mathbb R$ 

*There is a sequence of real numbers $\{x_{n}\}_{n=-\infty}^{n=\infty}$, and a constant $c>0$ such that $f(x_{n+1})-f(x_n)=c$ for all $n$.
(Edit: $\lim\limits_{n \to -\infty} x_{n}=-\infty$)
Now, is it true that $\lim \limits_{x \to -\infty} f'(x)$ cannot be zero?
I think it is true, because if $\lim\limits_{x \to -\infty} f'(x)=0$ then the slope of the tangent line at $-\infty$ will be very close (approach) to zero and this means--since $f$ is increasing--that $\lim\limits_{x \to -\infty}f(x)=a $, for some constant $a$, this can be seen by a graph!.
This means that the assumption is not correct because of item #2 above. Please correct me if my argument is not right!
Maybe my proof is not correct, but what about the problem itself! 
 A: Wrong, $\lim f'=0$ does not imply that $\lim f$ exists. Try $f(x)=x/\log(1+|x|)$.
Edit Try $f(x)=x^{17}/(1+x^{16}\log(1+|x|))$.

Maybe my proof is not correct, but what about the problem it self!

To answer that, let us consider the function $g:x\mapsto\sqrt{x}$ and limits when $x\to+\infty$. Then $g'(x)\to0$ but the limit of $g(x)$ does not exist, in fact $g(x)\to+\infty$. What the fact that the limit of $g'(x)$ exists and is zero for a nondecreasing $g$ implies, is not that $g$ has a limit but the (weaker) fact that the function $x\mapsto g(x)/x$ has a limit, and that this limit is zero. In the end the difference is partially similar to the difference between $o(x)$ and $O(1)$ when $x\to+\infty$.
A: Wouldn't $f(x) = -\ln(-x)$ work? The function is twice differentiable and for any $c>0\in\mathbb{R}$ we can find 
$$-\ln(-x_{n+1}) + \ln(-x_{n}) = c \iff x_{n} = e^{c}\cdot x_{n+1}$$
It's easy to see that this sequence satisfies the requirement
$$\lim_{n\rightarrow-\infty}x_{n} = -\infty$$
The function has derivative $-\frac{1}{x}$ so it is strictly increasing on its domain and
$$\lim_{x\rightarrow-\infty}f'(x) = \lim_{x\rightarrow-\infty}-\frac{1}{x} = 0$$
