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My question is about this question. We know that $f$ is differentiable on $(0,\infty)$ and $f(x)$ has a finite limit (let's say $L$) as $x\to\infty.$ My exact question is if we really need the assumption that $f'(x)$ has a finite limit as $x\to\infty.$

We can apply the mean value theorem to $f$ on the interval of the form $[n,n+1].$ Then, we have that there is a $x_n \in (n,n+1)$, such that: $$f'(x_n) = \frac{f(n+1) - f(n)}{(n+1) - n}= f(n+1) - f(n).$$ Now let $n\to \infty$. Then $\lim\limits_{n\to\infty} f(n+1) = \lim\limits_{n\to\infty} f(n) = L$, thus $\lim\limits_{n\to\infty}f'(x_n) = 0,$ with $x_n\overset{n\to\infty}{\longrightarrow}\infty.$

There is always the case I might miss something.

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thanasissdr
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  • it could be that $f'(x_n)$ does not exist – gt6989b Jul 01 '16 at 17:09
  • @gt6989b I edited a liitle bit my question. The function is differentiable on $(0,\infty).$ – thanasissdr Jul 01 '16 at 17:10
  • indeed, but that does not guarantee differentiability at $\infty$ – gt6989b Jul 01 '16 at 17:11
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    You have shown that there exists a number $n<x_n<n+1$ such that $f'(x_n)$ is arbitrarily small. But that does not imply that $\lim_{x\to \infty}f'(x)=0$. – Mark Viola Jul 01 '16 at 17:21
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    @Dr.MV So, actually I have shown there is at least one sequence $f'(x_n)$ that converges to zero, right? But since (in the original problem) $f'(x)$ has a finite limit as $x\to \infty$ , every convergent sequence $f'(a_n)$ will converge to the same limit, thus it is sufficient to find the limit of just one convergent sequence. Am I right? – thanasissdr Jul 02 '16 at 00:39
  • Yes, if the limit of $f'$ exists, then your method is valid. – Mark Viola Jul 02 '16 at 00:41
  • @Dr.MV Thank you for the insight. – thanasissdr Jul 02 '16 at 00:44

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Consider $$ f(x)=\frac{\sin\left(x^2\right)}{x} $$ then $$ \lim_{x\to\infty}f(x)=0 $$ yet $$ f'(x)=2\cos\left(x^2\right)-\frac{\sin\left(x^2\right)}{x^2} $$ which has no limit.

robjohn
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