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!
 A: Note that $f(x) \neq 0$ for all $x \geq 0$ shows that $f(x)$ must maintain a constant sign (if it changes sign then by intermediate value property it must vanish) for $x \geq 0$. Also the existence of $x_{0}$ with $f(x_{0}) > 0$ tells us that $f(x)$ is positive for $x \geq 0$. 
Next we are given that $\lim_{x \to \infty}f(x) = L$ exists. Since $f(x) > 0$ we must have $L \geq 0$. We will show that $L > 0$ leads to a contradiction. By the existence of limit $L = \lim_{x \to \infty}f(x)$ it follows that for any $\epsilon > 0$ there is a number $N > 0$ such that $$L - \epsilon < f(x) < L + \epsilon$$ for $x > N$. Choosing $\epsilon = L/2 > 0$ we see that the above implies that $f(x) > L / 2$ for $x > N$. Consider the function $f(x)$ in interval $[0, N]$. Clearly $f(x)$ is positive on this interval and hence its minimum value on this closed interval is also positive. Let this minimum value of $f(x)$ be denoted by $m$. Hence $f(x) \geq m$ for $x \in [0, N]$. And $f(x) > L / 2$ for $x > N$. It follows that $$f(x) \geq \min(m, L/2) = K (\text{say})$$ for all $x \geq 0$. We can now see that the condition (3) in the question is violated if we choose an $\epsilon$ with $0 < \epsilon < K$. Hence it follows that $L$ must be $0$.
A: 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.
A: 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 $\lim_{x\rightarrow \infty}f(x)=0$
A: 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!
