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$f:(0,\infty)\rightarrow\mathbb{R}$ is differentiable, $\lim_{x\rightarrow\infty}f(x)=1$, $\lim_{x\rightarrow\infty}f'(x)=c$ we need to show $c=0$

well, I tried like this $|f(x)-1|<\epsilon\forall x>M$, where $M$ is very large, $|f'(x)-c|<\epsilon\forall x>M$, what more I can say?thank you.

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up vote 3 down vote accepted

Since $\lim_{x \to \infty} \frac{f'(x)}{1}$ exists, by L'Hospital

$$\lim_{x \to \infty} \frac{f(x)}{x}=\lim_{x \to \infty} \frac{f'(x)}{1}=c \,.$$

But the first limit is $0$.

Alternately Use the MVT on the interval $[x,2x]$. You have


Now, when $x \to \infty$ we have $f'(c_x) \to f(c)$ and the LHS gos to $0$.

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Why can you use L'Hopital's like this? I thought you must have $\lim_{x \to \infty} f'(x)/g'(x)$ existing AND $f(x) \to \infty$ and $g(x) \to \infty$. – Antonio Vargas Jan 14 '13 at 16:53
@AntonioVargas Correct me if I am wrong, but in the proof of the $\infty/\infty$ case, we only use the fact that $g(x)$ goes to infty. – N. S. Jan 14 '13 at 18:42
For example, check the proof from Wiki: – N. S. Jan 14 '13 at 18:43
(+1) for this version of L'Hospital. – vesszabo Jan 14 '13 at 19:49

Yet another proof:

Let $0<x$. By the MVT in $(x,x+1)$: $$\exists \xi\in (x,x+1):f'(\xi)=\frac{f(x+1)-f(x)}{x+1-x}=f(x+1)-f(x)$$ Letting $x\to +\infty$, $\xi\to +\infty$ and so $$\lim_{x\to +\infty}f'(x)=1-1=0$$

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$\xi\to\infty$ ? – Un Chien Andalou Jun 17 '13 at 14:03

If c isn't 0 then f is increasing/decreasing and can't converge to a limit.

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Why? $\frac{\sin(x)}{x}$ has a limit at $\infty$.... – N. S. Jan 14 '13 at 16:50
@ N.S lim x-> inf (sinx/x)'=0. If f' does converge to a limit, it must be 0. – Ishan Banerjee Jan 14 '13 at 16:54
My point is that your reasoning is wrong. Namely this part:"f is increasing/decreasing and can't converge to a limit" is not true. – N. S. Jan 14 '13 at 18:38
Yeah,you're right. I meant monotonically increasing(kind of). if f' conveges to c>0 then for all x>a f'(x) >c-d for every d>0 so then we get f(x+b) >f(x)+b(c-d) if x>a aand hence f can't converge. I didn't really wan't to be rigorous. – Ishan Banerjee Jan 15 '13 at 7:41
$\arctan(x)$ is increasing but it converge ;) – N. S. Jan 15 '13 at 15:16

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