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!

First note that $f(x) >0$ for all $x$. This follows from 3. and the fact that $f$ is
continuous (and so $f([0,\infty))$ is connected). It follows from this that
$L \ge 0$.
Let $\alpha(x) = \min_{t \in [0,x]} f(t)$. Since $[0,x]$ is compact, we see that
$\alpha(x) >0$ for all $x$. Property 3. shows that $\lim_{x \to \infty} \alpha(x) = 0$.
If $L>0$, then there would be some $m>0$ such that $\alpha(x) \ge m$ for all $x$,
a contradiction.
