Let $f:\mathbb R \to \mathbb R$ be a continuously differentiable function such that $f(x)>0 , f'(x)>0 , \forall x \in \mathbb R$ ,

then is it true that $\lim _{x \to -\infty}f'(x)=0$ ?

I can only figure out that $\lim _{x \to -\infty} f(x)$ exists finitely as $f$ is increasing and bounded below .

Please help . Thanks in advance

  • 4
    $\begingroup$ No. Around the integer $-n$, make the graph of $f$ look like _/$\overline{\phantom{a} }$ with small height, smooth, and positive slope everywhere. $\endgroup$ – David Mitra Jul 16 '16 at 10:00
  • $\begingroup$ Sorry I misread the question, not seeing you wanted $f'$ not $f$ (I deleted my wrong example) $\endgroup$ – coffeemath Jul 16 '16 at 12:57

Just expanding the answer by David in the comment. Let $g$ be a positive continuous function so that

$$\int_{-\infty}^0 g(x) <\infty,\ \ \ g(-n) = n\ \ \forall n\in \mathbb N.$$


$$f(x) = \int_{-\infty}^x g(s) ds$$

satisfies your condition but $f' = g$ has no limit as $x\to -\infty$.

An example of such $g$ is

$$g(x) = \max\{ e^{x}, n(1- n^2| x+n|): n \in \mathbb N\}$$

  • $\begingroup$ I see , can you give a concrete example of such a $g$ ? $\endgroup$ – user228168 Jul 16 '16 at 10:45
  • $\begingroup$ Please see the edit. @SaunDev $\endgroup$ – user99914 Jul 16 '16 at 13:42
  • $\begingroup$ How do you show that $g(-n)=n$ for every n? $\endgroup$ – user84413 Jul 16 '16 at 21:40
  • $\begingroup$ @user84413 Note that $e^{-n} <1$ for all $n$, so $g(-n) = n(1-n^2|n-n|) = n$. $\endgroup$ – user99914 Jul 16 '16 at 21:45
  • 1
    $\begingroup$ Should $|x-n|$ be $|x+n|$ instead? $\endgroup$ – user84413 Jul 16 '16 at 22:47

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