# The function g:(0,infty)->R, is continuous, g(1)>0 and lim_(x->infty)g(x)=0 It is a fact that for every y between 0 and g(1) the function takes on a value in ( y, g(1) ) How would one show that if: lim_(e->0+)sup{g(x):0<x<e}=0

The function $g:\left(0,\mathrm{\infty }\right)\to \mathbb{R}$, is continuous, $g\left(1\right)>0$ and
$\underset{x\to \mathrm{\infty }}{lim}g\left(x\right)=0$
It is a fact that for every $y$ between 0 and $g\left(1\right)$ the function takes on a value in
How would one show that if:
$\underset{e\to 0+}{lim}sup\left\{g\left(x\right):0
then $g$ attains a maximum value on 0 to infinity. Does $g$ necessarily have to be bounded below for this to hold? Why?
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Hints
1. Since $\underset{x\to \mathrm{\infty }}{lim}g\left(x\right)=0$, there exists $R\in {\mathbb{R}}^{+}$ such that for $x>R$, $g\left(x\right).
2. Since $\underset{ϵ\to {0}^{+}}{lim}sup\left\{g\left(x\right):0, there exists $r\in {\mathbb{R}}^{+}$ such that for $x\in \left(0,r\right)$, $g\left(x\right).
3. $r<1.
4. The set $\left[r,R\right]$ is a compact set, and a continuous function on a compact domain achieves an absolute maximum on that domain. This is the Extreme Value Theorem.