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

  1. $f$ is an increasing function in $\mathbb R$

  2. 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!

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    Try $f=g^{-1}$, where $g(x)=x^3+x$. Or $f(x)=\operatorname{sgn}(x)\ln(|x|+1)$, where $\operatorname{sgn}(x)$ is $-1$ if $x<0$, $0$ if $x=0$, and $1$ if $x>0$.2011-11-01
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    the second function is not differentiable at x=0. What do you mean by $g^{-1}$, is it inverse or $1/g$?2011-11-01
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    It’s the inverse. The second function is differentiable at $0$: its derivative there is $1/2$.2011-11-01
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    Oops. Somehow I added $0$ and $1$ and got $2$. The derivative of the second function at $0$ is $1$.2011-11-01

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