"$\epsilon-\delta$" for $\lim_{x\to \infty}\sup_{\alpha\in [0,1]}f_{\alpha}(x)$ is bounded Suppose that I proved that $$\lim_{x\to \infty}\sup_{\alpha\in [0,1]}f_{\alpha}(x)\le C$$
where $C$ is a constant.
Does it means that there exist a constant $A$ such that $f_{\alpha}(x)\le C$ holds for every $\alpha\in[0,1]$ and $\vert x\vert\ge A$ ?
 A: Well given some $\beta\in [0,1]$, we have that for any $x$, $f_\beta (x) \leq \sup_{\alpha\in [0,1]} f_\alpha (x)$.  So in particular, it's true that 
$$\lim_{x\to\infty} f_\beta (x) \leq \lim_{x\to\infty} \sup_{\alpha\in [0,1]} f_\alpha (x) \leq C$$
for every $\beta\in [0,1]$.  We can write this as 
$$ \lim_{x\to\infty} (C - f_\beta (x) ) \geq 0 $$
By definition, we can take arbitrary $\epsilon >0$ and rephrase this as:
There exists an $A >0$ such that for all $|x| \geq A$, 
$$ C - f_\beta (x)  > -\epsilon $$
$$\implies  f_\beta (x)  < C + \epsilon $$
$$\implies f_\beta (x)  \leq C $$
A: Let $l$ be that limit and we fix $\epsilon>0$.
Then there exists $A$ such that for each $x\geq A$ we have 
$$
\sup_{\beta\in[0,1]}f_\beta(x)-l<\epsilon
.
$$
For an arbitrary $\alpha \in [0,1]$ we have
$$
f_\alpha(x)\leq \sup_{\beta\in[0,1]}f_\beta(x)<l+\epsilon
.
$$
Thus for each $\epsilon>0$ we have 
$f_\alpha(x)\leq l+\epsilon\to_{\epsilon\to 0^+}l\leq C$,
which means 
$$
f_\alpha(x)\leq C \quad\text{for each}\quad x\geq A
.\tag{1}
$$
Since $\alpha$ was chosen arbitrarily in $[0,1]$ equation $(1)$ holds for every $\alpha \in [0,1]$.

Concerning your question in the comment:
It is not possible to choose $A=1$ in general because, for example, regard
$$
f_\alpha(x)=f(x)=\chi_{[0,2]}
.
$$
Then you have that $l\leq C=\frac{1}{2}$ but for $|x|\geq 1$ it is not true that $f(x)\geq 1$ because, for example, if $x=1$, you have 
$$f(1)=1\not \leq \frac{1}{2}.$$
