Q: Continuous function with finite limit at infinite If $f:[0,\infty)→R$ a continuous function with a finite limit $\lim\limits_{x\to \infty}f(x)=L$ , then prove that $f$ is upper and lower bounded.
I am stuck at this last excercise and I can't think clearly since I am very tired. Every help is appreciated.
 A: Since $\lim_{x\to \infty} f(x)=L$ ,so given $\epsilon>0$ there exists $G$ (large) such that $x>G\implies |f(x)-L|<\epsilon\implies f(x)\in (L-\epsilon,L+\epsilon)$
Take $\epsilon =1\implies f(x)\in (L-\epsilon,L+\epsilon)$ forall $x>G$.
Now $f$ is continuous $[0,G]$ and a continuous function on a compact set is bounded so $f$ is bounded on $[0,G]$ 
So $f$ is bounded on both $[0,G] $ and $[G,\infty)$
A: Following Arthur's comment:
The fact that $\lim \limits_{x \to \infty} f(x) = L$ means given an arbitrary fixed $\epsilon > 0$, after some point $x^{*}$ we have that if $x \in (x^{*}, \infty)$, then $f(x)$ is in $(L - \epsilon, L + \epsilon)$.  This follows directly from the $\epsilon-\delta$ definition of a limit, so check that out.
So on the interval $[x^{*}, \infty)$, $f(x)$ is bounded from above by $L + \epsilon$ and from below by $L - \epsilon$.  Since $f$ is continuous, $f(x)$ is bounded on the closed interval $[0, x^{*}]$ by some number $M$ from above and by $m$ from below.  Let $s = \min\{m, L - \epsilon\}$ (i.e., $s$ is the minimum of the two lower bounds) and $S = \max\{M, L + \epsilon\}$ (i.e., $S$ is the maximum of the two upper bounds).  Then $s \leq f(x) \leq S$ for all $x \in [0,\infty)$.
