How to prove this limit is $0$? Let $f:[0,\infty )\rightarrow \mathbb{R}$ be a continuous function such that:  


*

*$\forall x\ge 0\:,\:f\left(x\right)\ne 0$.   

*$\lim _{x\to \infty }f\left(x\right)=L\:\in \mathbb{R}$.  

*$\forall \epsilon >0\:\exists x_0\in [0,\infty)$ that $0<f\left(x_0\right)<\epsilon $.   


Prove that: $$L=0$$
So far, I thought about assuming that $L\ne 0$ to get a contradiction.
Assume that $f\left(0\right)<0$. Now, choose an arbitrary $\epsilon $>$0$, and we know that $\exists x_0>0$ such that $0<f\left(x_0\right)<\epsilon$.
$f$ is continuous, so from the Intermediate value theorem, $\exists c\in \left[0,x_0\right]: \space f\left(c\right)=0$, and it's a contradiction to (1). So, $f\left(0\right)\ge 0$. (Actually, I think it means that $\forall c\in \left[0,\infty \right): \space f\left(c\right)\ge 0$, and it contradicts that $L<0$).  
I get stuck while proving the case when $L>0$. Can someone guide me? Thanks in advance!

Let $\{\epsilon_i\}$, ($\epsilon_i>0$, $i \in \mathbb{N}$) be  a sequence such that $\{\epsilon_i\} \rightarrow 0$.
From 3) , $\forall \epsilon_i$ we have a sequence $\{x_i\}$ such that $f(x_i)<\epsilon_i$ and a sequence $\{f(x_i )\} \rightarrow 0$. 
If it exists $\bar x$ such that all $x_i$ are in $[0,\bar x]$ then, from the  compactess of the interval, we musth have $\{x_i\} \rightarrow x_0\in [0,\bar x]$  and from continuity of the function $f(x_0)=\lim  \{f(x_i )\}=0$. That contradict 1). 
So, for any $\bar x$ there exists $\bar n$ such that $i>\bar n \Rightarrow x_i> \bar x$ and $f(x_i)<\epsilon$ and $\lim_{x\rightarrow \infty}f(x)=0$
