Finding supremum of a set; am I underdoing it? 
Find, with proof, the supremum of the subset $E$ of $\mathbb{R}$ defined by:
$E = \left\{\frac{25+16\sqrt{n}}{5+3\sqrt{n}}:n\in\mathbb{N}\right\}$

Is it ok to solve the limit of the function and call it a proof?
i.e. $$\text{sup}(E) = \lim_{n\to\infty} f(n) =\space ...$$
Or do I need to involve actual definitions of bounds etc.
 A: You have an increasing sequence
$$ \left\{\frac{25+16\sqrt{n}}{5+3\sqrt{n}}:n\in\mathbb{N}\right\}$$
with the limit as $n$ goes to infinity equal to $16/3$ 
The limit is the supremum, that is $$\sup (E) = 16/3$$
All you have to do is to show that the sequence is increasing and bounded above by its limit.
A: Hint
As I pointed out in my comment, in general, the limit may not be the answer for supremum. 
In this problem, observe that
$$\frac{25+16\sqrt{n}}{5+3\sqrt{n}}=5+\frac{1}{3+\frac{5}{\sqrt{n}}}.$$
So the given sequence is an increasing sequence (you still need to show boundedness, which is easy). So the limit will, give you the supremum. 
But simply computing the limit is not a proof that supremum is what you claim it to be.
A: The supremum is the limit if the function in question is increasing. Doing some reduction shows 
$$\frac{25+15\sqrt{n}}{5+3\sqrt{n}}=5+\frac{\sqrt{n}}{5+3\sqrt{n}}=
5+\frac{1}{
\frac{5}{\sqrt{n}}+3
}$$
which is clearly an increasing function of $n$. The supremum/limit of this function at natural number values is $5+1/3=16/3.$
