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$f(x)$ is a real valued function on the reals, and has a continuous derivative such that $$\lim_{x\to\infty} f'(x)^2 + f(x)^3 = 0.$$

How do i show that $$\lim_{x\to\infty} f(x) = \lim_{x\to\infty} f'(x)=0?$$

Note I tried to take $f(x)=\frac{1}{x}$ but it did not work

Thank you for any kind of help

gt6989b
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1 Answers1

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Suppose that : $\lim_{x\to\infty}f '(x)^2 + f(x)^3 = 0.$

  1. Show $\limsup_{x\to\infty} f(x) = 0:$ Showing $\,\limsup_{x\to\infty} f(x) \le 0$ is easy. If $\limsup_{x\to\infty} f(x) < 0,$ then $f(x) < -\epsilon$ for large $x$ for some $\epsilon > 0.$ Show this leads to a contradiction.

  2. If $\lim_{x\to\infty} f(x) = 0$ fails, then from 1., $\liminf_{x\to\infty} f(x) < 0.$ This shows there is $\epsilon > 0$ and sequences $x_1<y_1<x_2<y_2 < \cdots \to \infty $ such that $f(x_n)>-\epsilon /2,f(y_n) <-\epsilon $ for all $n$. Then for each $n$ $\min_{[x_n,x_{n+1}]} f = f(c_n)<-\epsilon $ for some $c_n\in (x_n,x_{n+1}).$ We then have $f'(c_n)=0$ and thus $f'(c_n)^2+f(c_n)^3$ does not approach $0,$ contradiction

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    Part 2 makes very little sense to me. – David C. Ullrich Jul 06 '15 at 18:03
  • you can use LH rule may it also help :$ \lim_{x\to\infty}f '(x)^2+f(x)^3 = 0 $ $\lim_{x\to\infty}\frac{{f(x)}^3}{{f'(x)}^2}=-1$ using LH rule...

    $\lim_{x\to\infty} \frac{3f(x)^2.f'(x)}{2f'(x).f''(x)}=\lim_{x\to\infty} \frac{3f(x)^2}{2f''(x)}=-1$ $\lim_{x\to\infty} \frac{6f(x).f'(x)}{2f'''(x)}=-1$

    – zeraoulia rafik Jul 06 '15 at 18:11
  • Ok, now I can read it. I still don't get it. For example I don't see why there exist $x_n$ and $y_n$ with the properties you claim. Not that it satifies the other conditions, but note that if f(x) = -1 for all $x$ then $\liminf f<0$ but there are no such $x_n$ and $y_n$. – David C. Ullrich Jul 06 '15 at 18:27
  • Regarding your comment, you should note that $a_n+b_n\to0$ does not imply $a_n/b_n\to-1$ (consider $a_n=1/n$, $b_n=1/n^2$. – David C. Ullrich Jul 06 '15 at 18:31
  • I said in my comment it's may work ,not sure – zeraoulia rafik Jul 06 '15 at 22:16