Prove that the limit of $2^{\frac{-1}{\sqrt{n}}}=1$ 
Prove that the limit of $2^{\frac{-1}{\sqrt{n}}}=1$.

I need to show that for each $\epsilon$ there exists an $n_0 \in \mathbb{N}$ such that $ \forall n \geq n_0: |2^{\frac{-1}{\sqrt{n}}}-1|\lt \epsilon$
I was simply trying to solve $2^{\frac{-1}{\sqrt{n}}}=1$ by taking logs, by that doesn't lead my anywhere. Can anybody help please?
 A: $\mathbf{Hint}$: Prove that the $\lim_{n\to \infty}\frac{-1}{\sqrt{n}}=0$. This should not be difficult seeing that you understand $\epsilon$-notation. 
A: No, it isn't good way, because from $2^{\frac{-1}{\sqrt{n}}}=1$ you get $\frac{-1}{\sqrt{n}}=0$-it's not possible.
What you can do:
1) If you know that for all $a>0$ $\lim_{n \to \infty} a^{\frac{1}{n}}=1$ you can use this.
2)If you don't know that for all $a>0$ $\lim_{n \to \infty} a^{\frac{1}{n}}=1$ you can prove this (there is, for example, proof using Bernoulli's inequality. 
A: Surely this works (first year undergrad here - please be gentle!)
$ 2^\frac{-1}{\sqrt n} -1 <\epsilon$
$ 2^\frac{-1}{\sqrt n} <\epsilon +1$
$ \log{2^\frac{-1}{\sqrt n}}<\log(\epsilon +1)$
$ \frac{\log 2}{\sqrt n} >-\log(\epsilon +1)  $
$ \frac{\sqrt n}{\log 2} < \frac{1}{\log(\epsilon +1)}  $
$ \sqrt n < \frac{\log 2}{\log(\epsilon +1)} $
$ n < \log(2-(\epsilon +1))^2 $
So that if you choose N $ \in N $ to be $\lceil \log(2-(\epsilon +1))^2 \rceil$ then $ \forall n >$N,  $|2^\frac{-1}{\sqrt n} -1| <\epsilon$.
