Showing a sequence is convergent using the ϵ-N definition I have just shown that the sequence
$\lim\limits_{n \to ∞}$ $( n-\sqrt{n^2-n} )$ = $\frac{1}{2}$
by using the difference of two squares. However I am having difficulty showing it converges to $\frac{1}{2}$ using the $\epsilon$-N definition of convergence. 
I start like this:
Let  $\epsilon$>0. Then let $N=\frac{1}{\epsilon}$. Then for all $n\geq N$, $|(n-\sqrt{n^2 -n}) -\frac{1}{2}| = ?= ...=\frac{1}{N} = \epsilon$
Therefore converges to $\frac{1}{2}$
I have been working through the khan academy videos on convergence but I cannot work out how their examples fit into mine and therefore fill my steps.
If anyone can explain my missing steps in detail I would be extremely grateful. 
Thanks in advance! 
 A: Note: For proofs like this you don't take $N$ to be something and try to prove that it works. You have to find the $N$ which work.
Why did you pick $N=\frac{1}{\epsilon}$? This $N$ does work, but the key to the proof is how you found it wot work.
Proof: You need
$$|(n-\sqrt{n^2 -n}) -\frac{1}{2}| <\epsilon \Leftrightarrow  \\
\left| \frac{n}{n+\sqrt{n^2-n}}-\frac{1}{2} \right| < \epsilon \Leftrightarrow  \\
\frac{1}{2}\left| \frac{n-\sqrt{n^2-n}}{n+\sqrt{n^2-n}} \right|< \epsilon \Leftrightarrow  \\
\frac{1}{2}\left| \frac{n}{(n+\sqrt{n^2-n})^2} \right| <\epsilon $$
Now use the fact that 
$$\frac{1}{2}\left| \frac{n}{(n+\sqrt{n^2-n})^2} \right| < \frac{n}{n^2}=\frac{1}{n}$$
Therefore, if you can make $\frac{1}{n} <\epsilon$ you get what you need.
Important: This is how we always reach the conclusion. Write 
$$|a_n - L| < \epsilon$$
and try to boil it down to something (which is implied by) 
$$n >\mbox{ something }$$
This tells you what your $N$ is..
Finally, you write the "nice" solution, which is your argument backwards:
Let $N$ be what you got at the last line, then write your algebraic manipulations backwards and get your stating line
$$|a_n-L|< \epsilon$$
A: $|n-\sqrt{n^2-n}-\frac{1}{2}|$
$=|\frac{n}{n+\sqrt{n^2-n}}-\frac{1}{2}|$
$=|\frac{1}{1+\sqrt{1-\frac{1}{n}}}-\frac{1}{2}|$
$=|\frac{1-\sqrt{1-\frac{1}{n}}}{2(1+\sqrt{1-\frac{1}{n}})}|$
$=|\frac{\frac{1}{n}}{2(1+\sqrt{1-\frac{1}{n}})^2}|$
$\leq|\frac{1}{n}|$
$\leq\varepsilon$
