Can I show that the limit of the square root is the square root of the limit this way? Given, for all natural k and a convergence sequence $a_n$:
$$
\lim_{n\to\infty}\left((a_n)^k\right)=\left(\lim_{n\to\infty}a_n\right)^k
$$
I have to show that $\lim_{n\to\infty}\sqrt{1-\frac{2}{n}}=1$. So:
$$
(a_n)^2 = 1-\frac{2}{n}=> \lim_{n\to\infty}\left((a_n)^2\right)=\lim_{n\to\infty}\left(1-\frac{2}{n}\right)=1=\lim_{n\to\infty}\sqrt{1-\frac{2}{n}}
$$
Can I do this? thank you.
 A: You can't use that method directly, because you haven't yet shown that $a_n$ converges. Fortunately, you can say first that $a_n < 1$ for all $n$, and it is an increasing sequence, so must converge. Then, your argument works.
Alternatively (in my opinion the more natural approach), observe that $1 - \frac{2}{n} < \sqrt{1 - \frac{2}{n}}< 1$, so we have convergence by the squeeze theorem.
Specifically, 
$$\begin{align} \sqrt{1 - \frac{2}{n}} < 1 \implies \sqrt{1 - \frac{2}{n}}\times \sqrt{1 - \frac{2}{n}} < \sqrt{1 - \frac{2}{n}} \implies 1 - \frac{2}{n} < \sqrt{1 - \frac{2}{n}}\end{align}$$
A: Note that,
if $0 < x < 1$,
then
$(1-x)^2 = 1-2x+x^2
=1-x-x+x^2
=1-x-x(1-x)
\lt 1-x
$,
so,
taking square roots,
$1-x < \sqrt{1-x}
$.
This lets you squeeze.
If you want an upper bound,
$(1-x/2)^2
=1-x+x^2
\gt 1-x
$
so
$1-x/2 > \sqrt{1-x}$.
Therefore,
if $0 < x < 1$,
$1-x < \sqrt{1-x}
\lt 1-x/2$.
A: If worse comes to worse we can do it directly.
For $\epsilon > 0$ we want to show there is an $N_\epsilon$ so that whenever $n > N_\epsilon$ we have $|\sqrt{(1 - \frac 2n)} - 1|< \epsilon$.
Let's work backwards.
For $2< n$ then $\frac 2n > 0$ so $1-\frac 2n < 1$ so $\sqrt{(1-\frac 2n)} < 1$ so $|\sqrt{1-\frac 2n)} -1| = 1- \sqrt{1-\frac 2n} < \epsilon \Leftarrow$ ("is implied by"; we are working backwards).
$1-\epsilon < \sqrt{1-\frac 2n}\Leftarrow$
$(1-\epsilon)^2 = 1- 2\epsilon + \epsilon^2 < 1-\frac 2n$
As we may assume $\epsilon < 1$ so $-2\epsilon + \epsilon^2 < -2\epsilon + \epsilon = - \epsilon$$
$\Leftarrow 1- \epsilon < (1-\epsilon)^2 < 1 - \frac 2n \Leftarrow$
$\frac 2n < \epsilon; \epsilon < 1\Leftarrow$
$\frac 2\epsilon < n$ so let $N = \frac 2\epsilon$ and... we are done.
....
Of course, no one likes to do that and we invented squeeze theorems and analysis of continuous functions, etc. just so that we don't HAVE to keep doing this stuff.  
But... If worse comes worse, we always can.
