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Let $f : (0,+\infty) \rightarrow (0,+\infty)$ and $g$ being a derivative of $f$ such as $g(x)=f^2(x)+f(x), \forall x>0$. Prove that $\lim_{x \to \infty} \frac{f(x)}{x} = 1/2$

John
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    What do you mean by $f^2(x)$? And when you say "$g$ is a differential of $f$", do you mean $f'=g$? – Jack M Aug 08 '14 at 20:03
  • It is a textbook question but I am quite certain that it means $(f(x))^2$, and it says that "there is a $g$ being a differential of $f$" so I guess it could be any derivative. – John Aug 08 '14 at 20:04
  • It's still not at all clear what the question means. "A differential of f" does not at all directly mean "$g(x)=(f(x))^2$". And with no condition at all on $g$, what's that relation giving on $f$? Something's mis-copied... – paul garrett Aug 08 '14 at 20:08
  • I am trying to translate it correctly, messed up differential and derivative. I honestly thought it was the same thing. – John Aug 08 '14 at 20:11
  • @John so you're saying that $f'(x) = (f(x))^2 + f(x)$? – DanZimm Aug 08 '14 at 20:12
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    It is not clear to me either so I assumed that it could be any $n$ such as the $n$-th derivative of $f$ is $g$. If it was the first derivative I see no reason that the author didn't write $f'(x)=f^2(x)+f(x)$. – John Aug 08 '14 at 20:14

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The solutions of this equation are of the form $\dfrac{\lambda e^{x}}{1-\lambda e^x}$ with $\lambda<0$, is not working.

Hamou
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