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If f→L as x→∞ and f is of class C1(f'is continuous) Does it imply that f'→0 as x→∞?? enter image description here the image is from solution manual of Advanced Engineering Mathematics, 5th edition, Kreyszig, Chapter2.1 exercise34.

breakgood
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  • This might be helpful: http://math.stackexchange.com/questions/1038951/continuous-increasing-bounded-function-derivative/2185440#2185440 – zhw. Mar 15 '17 at 17:22

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In general, this isn't true. For example, the function $$y(x) = \frac{\sin(x^{10})}{x}$$ has limit zero as $x\to\infty$ but $$y'(x) = \frac{10x^9\cos(x^{10})}{x} - \frac{\sin(x^{10})}{x^2}$$ does not have a limit as $x \to \infty$. Perhaps the author is invoking some other property of $y$ without explicitly stating it.

User8128
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  • It is mentioned that $y(x)$ is increasing. – Hans Lundmark Mar 15 '17 at 14:59
  • Even when $f$ is increasing, you can't guarantee existence of the limit of $y'$ at $\infty$. Let $u$ be a function which has smooth humps on the intervals $(n - 1/n^3,n+1/n^3)$ and is zero elsewhere. The humps can be very tall (height on the order of $n$) and the area underneath will be on the order of $1/n^2$. Then $y(x) = \int_0^x u(t)dt$ will have a finite limit at $\infty$ by $y'(x) = u(x)$ has no limit at $\infty$. – User8128 Mar 15 '17 at 21:57
  • Ah, yes, that's true... But maybe $y'(x)$ needs to be unbounded then? So it would contradict $y'=f(y)$ where $y$ is bounded and $f$ is continuous? Anyway, there seems to be something missing in the argument in the solution manual. – Hans Lundmark Mar 16 '17 at 07:33
  • If I'm guessing correctly, the exercise seems to be to prove that if $f(y)=0$ for $y=c_1$ and $y=c_2$, and $f(y)>0$ in between, then a solution to $y'=f(y)$ which starts between $c_1$ and $c_2$ must tend to $c_2$ as $x \to \infty$. As argued in the solution manual, $y(x)\to L$ for some $L\le c_2$. But then one can argue as follows: if $L<c_2$, then $y'(x) \to f(L)>0$, so there is $\omega$ such that $y'(x) > f(L)/2 > 0$ for all $x>\omega$, forcing $y(x) \to \infty$, contradiction. Hence $L=c_2$. – Hans Lundmark Mar 16 '17 at 07:37
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$f:\mathbb{R}\to\mathbb{R}$ defined by :

$$\forall x\in\mathbb{R},\,f(x)=e^{-x}\sin(e^x)$$

$f$ is $C^\infty$ and $\lim_{+\infty}f=0$, but $f'$ doesn't have any limit at $+\infty$.

Adren
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