Proving that the limit of the following sequence $\frac {\sqrt {n}}{1 + \sqrt{n}}$ when $n \rightarrow \infty$ is 1. Proving that the limit of the following sequence:
$$\lim_{n \to \infty} \frac {\sqrt {n}}{1 + \sqrt{n}} = 1$$
I have selected the $N$ to be $\lfloor (\frac{1}{\epsilon} - 1)^2 \rfloor + 1$, am I right?
 A: We have
$$ \left|\frac{\sqrt{n}}{1+\sqrt{n}} - 1\right| = \frac{1}{1+\sqrt{n}} , $$
so we want
$$ \frac{1}{1+\sqrt n} < \epsilon \iff \sqrt{n} > \frac{1}{\epsilon}-1 \iff n > \left( \frac{1}{\epsilon}-1 \right)^2, $$
so you can indeed take
$$ N = \left\lfloor \left( \frac{1}{\epsilon}-1 \right)^2 \right\rfloor + 1. $$
A: I think you have it.
We know $0 < \sqrt n < 1+ \sqrt n$ so $\frac {\sqrt n}{1 + \sqrt n} <1$ so we need to prove that for any $\epsilon > 0$ then there is an $N$ so the $n > N$ would imply
$0< 1 - \frac {\sqrt n}{1 + \sqrt n} < \epsilon$
And $1 - \frac {\sqrt n}{1 + \sqrt n} < \epsilon \iff$
$(1 + \sqrt n) - \sqrt n= 1 < \epsilon(1+\sqrt n) \iff $
$\frac 1{\epsilon} -1 < \sqrt n$.  Wolog we may assume $\epsilon < 1$. If we prove it is true for any $0 < \epsilon_1 < 1$ then the same $N$ will work for any larger $\epsilon \ge \epsilon_1$.  So if we assumme $0 < \epsilon < 1$ then $\frac 1{\epsilon} > 1$ and
$\frac 1{\epsilon} -1 < \sqrt n \iff$
$(\frac 1{\epsilon} -1)^2 < n$ so
So for any $N \ge (\frac 1{\epsilon} -1)^2$ we will have that if $n > N$ then $0< 1 - \frac {\sqrt n}{1 + \sqrt n} < \epsilon$.
So, ... Okay, I'd have picked the ceiling of $(\frac 1{\epsilon} -1)^2$ and there's no brownie points in figuring the exact minimum $N$ we can choose.
But yes if $N = \lfloor (\frac 1{\epsilon} -1)^2 \rfloor < n$ and $n \in \mathbb N$ then $n \ge N+1 >  (\frac 1{\epsilon} -1)^2$ and therefore $0< 1 - \frac {\sqrt n}{1 + \sqrt n} < \epsilon$.
So yes.  Using the floor will work and is correct.  And so will any $N$ greater than the floor.
Actually, is there any requirement in your definition that $N$ must actually be a natural number?  There is no logistical requirement for this if you think about it.  After all it is $n > N$ that must be natural.   The $N$ is just a hurdle and there's no reason that can't just be some real magnitude.
So actually, I'd just choose $N = (\frac 1{\epsilon} -1)^2$ and be done with.
A: As 
$$\frac{\sqrt n}{1+\sqrt n}-1=\frac1{1+\sqrt n},$$ which is a decreasing function, you can work this out by solving the equation
$$\frac1{1+\sqrt x}=\epsilon,$$ the solution of which is obviously
$$x=\left(\frac1\epsilon-1\right)^2.$$
Now you want the first integer $n$ strictly larger than $x$, and this is $\lfloor x\rfloor+1$.
A: Let $f(n)$
be a increasing
positive sequence
such that,
for any $c > 0$,
there is an $N(c)$
such that
$f(N(c)) > c$.
Then,
for any reals $a$ and $b$,
$\lim_{n \to \infty} \dfrac{f(n)+a}{f(n)+b}
=1$.
Proof.
$\dfrac{f(n)+a}{f(n)+b}-1
=\dfrac{(f(n)+a)-(f(n)+b)}{f(n)+b}
=\dfrac{a-b}{f(n)+b}
$.
For any $\epsilon > 0$
we want an $n$ such that
$|\dfrac{a-b}{f(n)+b}|
\le \epsilon$.
This will be true if
$\dfrac{|a-b|}{f(n)-|b|}
\le \epsilon
$
or
$|a-b|
\le \epsilon(f(n)-|b|)
$
or
$f(n)
\ge \dfrac{|a-b|}{\epsilon}+|b|
$.
Therefore,
if 
$n > N(\dfrac{|a-b|}{\epsilon}+|b|)
$,
$|\dfrac{f(n)+a}{f(n)+b}-1|
\lt \epsilon
$,
so
$\lim_{n \to \infty} \dfrac{f(n)+a}{f(n)+b}
= 1$.
Now let
$a=0, b=1,
f(n) = \sqrt{n}$.
We can choose
$N(c) = c^2+1$.
