# Limit infimum argument

Is the following true?

If $\liminf_{x\to\infty}f(x)>c$, where $c>0$,

then:

given $\epsilon >0$ there exist $x_{\epsilon}>0$ such that $f(x)>c+\epsilon$ , for all $x\geq x_{\epsilon}$.

The thing that I know is that: there exist $x_{o}>0$ such that $f(x)>c$ , for all $x\geq x_{o}$, (so no $\epsilon$ involved!)

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We have that $l:=\lim_{x\to +\infty}\inf_{t\geq x}f(t)>c$. So, given $\varepsilon>0$ (small enough, not all), we can find $x_0$ such that if $x\geq x_0$ then $\inf_{t\geq x}f(t)\geq c+2\varepsilon$. So, if $x\geq x_0$, $f(x)\geq \inf_{t\geq x}f(t)\geq c+2\varepsilon>c+\varepsilon$.

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The statement you are asking about is only true if $\epsilon$ is small enough and in particular $$\epsilon < \liminf_{x\to\infty}f(x) - c$$

For example if $f(x)=42$, $c=41.9$ and $\epsilon =0.5$ then $f(x) \lt c + \epsilon$ for all $x$.

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So do you mean that the statement would be true if we say: There exists $\epsilon_{o}>0$ and $x_{o}$ such that $f(x)>c+\epsilon_{o}$, for all $x\geq x_{o}$.!! – Marchael Apr 22 '12 at 10:31
Yes - or you could say "there exists $\epsilon_0 \gt 0$ such that for any $\epsilon$ with $\epsilon_0 \gt \epsilon \gt 0$ there exists $x_{\epsilon}$ such that $f(x)>c+\epsilon$ for all $x\geq x_{\epsilon}$". – Henry Apr 22 '12 at 10:38
Thank you. But how I can prove this? Any reference! – Marchael Apr 22 '12 at 10:58
If you know $\epsilon < \liminf f(x) - c$ then apply "the thing that I know" in italics to $c+\epsilon$ – Henry Apr 22 '12 at 11:53
Ok. Now just to summarize things: If $\liminf_{x\to\infty} f(x)>c>0$, then we can find $\epsilon_{o}>0$ small enough (this is always true!) so that $\liminf_{x\to\infty} f(x)-c>\epsilon_{o}$. As a consequence, we will have the following: for any $\epsilon$ with $0<\epsilon<\epsilon_{o}$, there exists $x_{\epsilon}$ such that $f(x)>c+\epsilon$ for all $x\geq x_{\epsilon}$. Plesae correct me if I misunderstand! – Marchael Apr 22 '12 at 18:36