Limit infimum argument

Question

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 $x\geq x_{\epsilon}$.

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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!)

Answer 1

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 $x\geq x_0$, $f(x)\geq \inf_{t\geq x}f(t)\geq c+2\varepsilon>c+\varepsilon$.

Answer 2

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 $x$.