Verification of proof concerning the supremum of a set. Given some set say $A$ = {$ x\over(x+7)$|$x \in \mathbb{R^+}$} 
Where 
$x\over(x+7)$ = 1 - $7\over(x+7)$ $\lt 1$  
Hence $1$ is an upper bound of $A$.
Method $1$. 
Consider $SupA = \alpha$ where  $\alpha \lt 1$
$x \ge 0$   for all $x \in \mathbb{R^+}$
$\therefore$  $x+7\ge7$
$\therefore$ $(1-\alpha)(x+7)\gt7$
$\therefore 1-\alpha\gt$ $7\over(x+7)$
$\therefore -\alpha\gt$$-1 + $$7\over(x+7)$
$\therefore \alpha \lt 1 - $$7\over(x+7)$ 
Hence $\alpha \lt $$x\over(x+7)$
Therefore $\alpha$ is not an upper bound of $A$ and hence $SupA = 1$
Method $2.$
For $SupA = 1$ to be true the following must hold for any arbitrary $\epsilon \gt 0$
$ 1 - \epsilon \lt A^{'}$ for some $A^{'} \in A$
i.e. $1 - \epsilon \lt $$x\over(x+7)$ for some $x\in \mathbb{R^+}$
Take $x+7 \gt $$7\over \epsilon$
$\therefore \epsilon \gt $$7\over(x+7)$
$\therefore -\epsilon \lt $-$7\over(x+7)$
$\therefore 1-\epsilon \lt 1-$$7\over(x+7)$
Hence $1-\epsilon \lt $$x\over(x+7)$ for some $x \in \mathbb{R^+}$
Therefore $SupA = 1$
Are either of these methods of proving supremum correct? If both, which one is better? If neither, how do i do it? 
Note: please excuse poor formatting, haven't learnt how to do it yet
 A: The second method looks fine.
The first method is incorrect. You wrote $x+7\ge 7$ and then $$(1-\alpha)(x+7)>7$$ That's your mistake.
Remark: If you know how the $\sup$ is characterized by limits of sequences, then you can easily show your result using the fact that $$\lim_{n\to +\infty}\frac{n}{n+7}=1$$
Edit For your curiosity: If $A$ is an upper-bounded non-empty set and $s\in\mathbb{R}$, to show that $s=\sup A$ you can simply show that $s$ is an upper-bound of $A$ and that there exists a sequence $(a_n)_{n\in\mathbb{N}}$ of elements in $A$ (i.e $\forall n\in\mathbb{N},a_n\in A$) such that $\lim\limits_{n\to +\infty}a_n=s$. That's because if such a sequence exists, it means that $$\forall\varepsilon>0,\exists N\in\mathbb{N},\forall n\ge N,s-\varepsilon<a_n\le 1$$ and in particular $s-\varepsilon<a_N$ with $a_N\in A$. In fact, you can also prove that if $s=\sup{A}$ then such a sequence exists.
A: Your first method has a big flaw: from $x+7\ge7$ it doesn't follow that
$$
(1-\alpha)(x+7)\ge7
$$
For instance, if $\alpha=1/2$ and $x=1$,
$$
(1-\alpha)(x+7)=\frac{1}{2}\cdot8=4<7
$$
Indeed, with this false statement, you conclude that $\alpha<x/(x+7)$ for every $x\in\mathbb{R}^+$, which is certainly false when $0<\alpha<1$.
What you want to prove is that $\alpha<x/(x+7)$ for (at least) one $x\in\mathbb{R}^+$. The inequality is equivalent to $\alpha<1-\frac{7}{x+7}$, that is
$$
\frac{7}{x+7}<1-\alpha
$$
or
$$
x>\frac{7}{1-\alpha}-7=\frac{7\alpha}{1-\alpha}
$$
which is certainly satisfied for some (actually infinitely many) $x\in\mathbb{R}^+$.
The second method is essentially the same: for $\varepsilon>0$ you want to see that there exists $x\in\mathbb{R}^+$ such that
$$
1-\varepsilon<\frac{x}{x+7}
$$
that is
$$
\varepsilon>\frac{7}{x+7}
$$
that's equivalent to
$$
x>\frac{7}{\varepsilon}-7
$$
Thus, taking $x=7/\varepsilon$ you are done.
There is also a third method, in this case.
Clearly, any $\alpha\le0$ is not an upper bound for the set $A$. Assume $0<\alpha<1$: we show that
$$
\frac{1+\alpha}{2}\in A
$$
Indeed
$$
\frac{1+\alpha}{2}=\frac{x}{x+7}
$$
becomes
$$
(1+\alpha)x+7(1+\alpha)=2x
$$
that is
$$
x=\frac{7(1+\alpha)}{1-\alpha}\in\mathbb{R}^+
$$
Since $\alpha<\frac{1+\alpha}{2}$, we conclude $\alpha$ is not an upper bound for $A$.
