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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 $x>0$. – Brian M. Scott Nov 1 '11 at 12:12
the second function is not differentiable at x=0. What do you mean by $g^{-1}$, is it inverse or $1/g$? – Kristen Nov 1 '11 at 12:27
It’s the inverse. The second function is differentiable at $0$: its derivative there is $1/2$. – Brian M. Scott Nov 1 '11 at 12:36
Oops. Somehow I added $0$ and $1$ and got $2$. The derivative of the second function at $0$ is $1$. – Brian M. Scott Nov 1 '11 at 19:02

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 $x\to+\infty$.

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your example is not differentiable at x=0. – Kristen Nov 1 '11 at 12:23
Do you think it matters as regards the behaviour at $-\infty$? – Did Nov 1 '11 at 12:50
I don't think using these examples one can easily verify the second property above! Of course there are many functions with limit of $f'$ at $-\infty$ is zero and the limit of $f$ doesnot exist, but I have here a very important property that $f$ has, item #2! I think it should play some role in this context? – Kristen Nov 1 '11 at 13:04
Wrong, the second example (the one with the exponent 17) does. Item #2 (for continuous nondecreasing functions) is just a complicated way to say that the limit of $f$ at $-\infty$ is $-\infty$ and the limit of $f$ at $+\infty$ is $+\infty$ (simply define $x_n$ by $f(x_n)=n$, for every integer $n$, and choose $c=1$). You could try to see why the exponent-17-example works. – Did Nov 1 '11 at 13:09

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 $$\lim_{x\rightarrow-\infty}f'(x) = \lim_{x\rightarrow-\infty}-\frac{1}{x} = 0$$

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The example shows what’s wrong with Kristen’s argument, but it appears that she wants a function whose domain is $\mathbb{R}$. – Brian M. Scott Nov 1 '11 at 19:05

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