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!

1)By condition 3 we can find a sequence $(x_n)$ such that $\lim_{n\to\infty} f(x_n)=0$. 
2)The sequence $(x_n)$ cannot be bounded(because otherwise by picking a proper subsequence , using 1. and continuity we conclude that $f(t)=0$ for some $t$) , 
3)By taking a subsequence $x_m\to \infty$ as $m\to\infty$,if necessary, we find that $\lim_{m\to\infty} f(x_m)=0$.so So $\lim_{x_m\to\infty} f(x_m)=0$.  This together with 3. implies $L=0$.
Remark:we need condition 1. Consider $f(x) = 1-\frac{1}{1+x^2}$ it satisfies 2 and 3 but not 1.conclusion doesn't follow!
