x \stackrel{x \to \infty}{\longrightarrow} 0 $.

Asked Feb 08 '13 at 18:03

Active Feb 09 '13 at 12:49

Viewed 93 times

First-time poster. Please forgive me if I do something unorthodox.

To be more specific, I need a function $f:[0,+\infty)\to[0,+\infty)$ so that $f$ is nondegenerate, nondecreasing, continuous, $$\lim_{x\to+\infty}\frac{e^{f(x)}}{x}=0$$ and $\forall\,\varepsilon>0\,\exists\,x_0\geq0$ such that $$0<-\left(f'(x)+\frac{f''(x)}{f'(x)}\right)<\frac{\varepsilon}{x}$$ for $x>x_0$.

I've tried a good number of functions and I just can't get it to work. I'm beginning to think there may be no such function.

Also, it would be fine if $f$ is not an elementary function. Like start with $f'$ and get an antiderivative.

Thanks in advance.

edited Feb 09 '13 at 12:49

asked Feb 08 '13 at 18:03

danzibr

1

what is a nondegenerate function? – Maesumi Feb 08 '13 at 18:54
What you'd expect: $f(x)=0\iff x=0$. – danzibr Feb 08 '13 at 20:03

Need an $ f $ so that $ 0 < -\left(f'(x) + \frac{f''(x)}{f'(x)}\right) < \epsilon/x $ and $ e^{f(x)}/x \stackrel{x \to \infty}{\longrightarrow} 0 $.

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