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Is it possible to have $$\lim_{x\to+\infty}x\left[f(x)\right]^5=0$$ and $$\lim_{x\to+\infty}x\left[f(x)\right]^3=+\infty$$ for some function $f$ at the same time?

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Take $f(x)=x^{-\frac{1}{4}}$ – gammatester Jun 6 '14 at 10:00
Does $f^3(x)$ mean $[f(x)]^3$ or $f(f(f(x)))$? – JiK Jun 6 '14 at 10:00
up vote 11 down vote accepted

Hint: Consider $x\mapsto x^\alpha$, for an appropriate $\alpha$.

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Yes. For example, $f(x)=\dfrac{1}{\sqrt[4]x}$

$\lim\limits_{x\to\infty} \dfrac{x}{(\sqrt[4]{x})^3}=\infty$

$\lim\limits_{x\to\infty} \dfrac{x}{(\sqrt[4]x)^5}=0$

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If you look at $$L(n)=\lim_{x\to+\infty}xf^n(x)$$ as Git Gud suggested, use $f(x)=x^a$ and so you search for $$L(n)=\lim_{x\to+\infty}x^{an+1}$$ So, if $an+1 \gt 0$, the limit is $\infty$ and if $an+1 \lt 0$, the limit is $0$.

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Assume $f(x) = x^P$, then to satisfy conditions listed,

  1. $x^{5P + 1} \rightarrow 0 \Rightarrow 5P + 1 < 0$
  2. $x^{3P + 1} \rightarrow \infty \Rightarrow 3P + 1 > 0$

So you get: $f(x) = x^P$, where $P \in \left]-\dfrac{1}{3}, -\dfrac{1}{5}\right[$

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