Showing a function is limited This is something about which I'm quite confident, but I would really like to be sure
Let $f(x)$ be a continuos function in $\mathbb{R}$
Then if $\displaystyle \lim_{x \to \pm\infty} f(x) = l \in \mathbb{R}$, then $f(x)$ is bounded. (There exists $a, b$ such that $a \le f(x) \le b \ \forall x \in \mathbb{R}$)
I think I can come up with a formal proof, but a more naive one would be to say that in any finite intervals there exist such $a, b$ (it's just the weirstrass theorem).
If trouble happens, it has to be in a neighborood of infinity.
So if the limit is finite, this means that also there the function is bounded above (not necessarily by $l$) so it's proven
Any feedback appreciated
Formal Proof
Let's focus on the limit to $+\infty$. 
$\displaystyle \lim_{x \to \pm\infty} f(x) = l \Rightarrow$
$$\forall \varepsilon > 0 \ \exists N : \forall x \ge N : |f(x) - l| < \varepsilon$$
This means that $\forall \varepsilon$, calling $\displaystyle M_{\epsilon} = \max_{x \in (0, N)} f$    we can say that $f(x) \le \max(M_{\epsilon}, l + \epsilon)$
Since $M_\epsilon$ is always finite, and so is $l + \epsilon$, we have a proof.
 A: Your proof is unusual, it shows an entire family of bounds. You only consider $x > 0$, and you only produce upper bounds on $f(x)$, no lower bounds. You know how to address these points, however, and provide lower bounds for the values of $f$, and bounds for the values at negative $x$, I suppose.
There's a technical detail wrong, you wrote
$$\max_{x\in (0,N)} f,$$
but for the open interval $(0,N)$, the maximum need not exist, so you should use the closed interval $[0,N]$ there (or write $\sup$).
But then it works.

The usual way to prove the boundedness of $f$ would be to choose one particular $\varepsilon > 0$. $\varepsilon = 1$ is a common choice. Then, since
$$\lim_{x\to+\infty} f(x) = l_+,$$
there is an $N\in\mathbb{R}$ with $x\geqslant N \Rightarrow \lvert f(x)-l_+\rvert \leqslant 1$, so $\lvert f(x)\rvert \leqslant \lvert l_+\rvert + 1$ on $[N,+\infty)$. Since
$$\lim_{x\to-\infty} f(x) = l_-$$
(we have no reason to demand $l_+ = l_-$, so let's do it for the more general situation), there is an $M\in \mathbb{R}$ with $x \leqslant M \Rightarrow \lvert f(x)-l_-\rvert \leqslant 1$, so $\lvert f(x)\rvert \leqslant \lvert l_-\rvert+1$ on $(-\infty,M]$.
If $M < N$, let $A = \max\limits_{x\in [M,N]} \lvert f(x)\rvert$, else let $A = 0$. Then we have
$$\lvert f(x)\rvert \leqslant \max \{ A ,\, \lvert l_-\rvert + 1,\, \lvert l_+\rvert + 1\}.$$
