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Let $f\colon\mathbb{R}\to\mathbb{R}^n$ be a $C^\infty$ function such that $\lim_{t\to\pm\infty}f^{(k)}(t)$ exists for all $k\in\mathbb{N}\cup\{0\}$. A neat little trick (apparently due to Thomas Hill) shows that $\lim_{t\to\pm\infty}f^{(k)}=0$ for $k\geq 1$. Indeed by l'Hopital's rule $$\lim_{t\to\pm\infty}f(t) = \lim_{t\to\pm\infty}\frac{e^t f(t)}{e^t} = \lim_{t\to\pm\infty}\frac{e^t(f(t)+f'(t))}{e^t}=\lim_{t\to\pm\infty}(f(t)+f'(t)),$$ so the claim follows inductively.

Is it possible to draw conclusions about the asymptotic decay of $f'$? For example, is there a $\delta > 0$ such that $\lim_{t\to\pm\infty}f'(t)\cdot e^{\delta |t|}=0$? Are there examples when this is not the case?

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What about $f(x) = \dfrac{1}{1+x^2}$? – Siméon Jan 17 '13 at 13:26
@Ju'x that looks like an answer to me. Should be fairly easy to prove by induction that $f^{(n)}(x)$ has the form $p(x)/q(x)$ where $q(x)$ has degree $2n+2$ and $p(x)$ has degree $n-1$. – JSchlather Jan 17 '13 at 13:52
Yeah, it answers this part of the question. Still is there anything that can be said about the asymptotic decay of f'? – jds Jan 17 '13 at 13:59
@jds, What else do you want to know? The limit your curious about goes to infinity in Ju'x's example. – JSchlather Jan 17 '13 at 14:32
@Jacob Schlather, A nice answer would be that there always exists a $C^\infty$ function $\phi\colon\mathbb{R}\to\mathbb{R}$ with $\lim_{t\to\pm\infty} \phi(t)=\infty$ such that $\lim_{t\to\infty}f'(t)\phi(t)=0$. I'm rather new on this website, should I edit/amend the question? – jds Jan 17 '13 at 14:42

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