Elementary proof for $\lim_{n\to \infty}\sqrt[2n-1]{\dfrac{1}{2^n}}=\dfrac1{\sqrt 2}$ This question is from Rudin PMA Ch 2:

How to show that $\lim_{n\to \infty}\sqrt[2n-1]{\dfrac{1}{2^n}}=\dfrac1{\sqrt 2}$? 

It was pointed out to me that $\lim_{n\to\infty}\dfrac1{2^{\frac n{2n-1}}}=\dfrac1{2^{\lim_{n\to\infty}\frac n{2n-1}}}$, but I got to know that 'taking limit to power' depends on continuity. Rudin has yet to touch continuity. 
Can we use the fact that $\lim_{n\to \infty}\sqrt[2n]{\dfrac{1}{2^n}}=\dfrac1{\sqrt 2}$ for this  proof?
 A: With $m=2n-1$, the expression is
$$\left(2^{-(m+1)/2}\right)^{1/m}=2^{-1/2}2^{-1/2m}.$$
Clearly the second factor tends to $1$.

If this needs to be proven, let $2^{1/2m}=1+\epsilon>1$. Then by the binomial theorem,
$$(1+\epsilon)^{2m}=2=1+2m\epsilon+\cdots$$ (all terms are positive) and $2m\epsilon<1$.
Hence
$$1<2^{1/2m}<1+\frac1{2m}$$ which squeezes to $1$.
A: You can use the fact that $\lim_{n\to\infty}\sqrt[2n]{\frac{1}{2^n}} = \frac{1}{\sqrt 2}$ as follows. For each $n$, let $\delta_n > 0$ be such that
$$
\delta_n + \frac{n}{2n} = \frac{n}{2n-1}.
$$
Note that $\lim_{n\to\infty}\delta_n = \lim_{n\to\infty}\frac{1}{2(2n-1)}= 0$ by the Archimedean property of $\Bbb R$ and that $\lim_{n\to\infty}2^{-\delta_n}=\sup_{r<0} 2^r = 2^0 = 1$ where the sup is over all rationals $r < 0$. Note that this doesn't use continuity, just properties of suprema and the fact that $\delta_n\to 0$: If $\epsilon > 0$, then for all sufficiently large $n$, we have $-\delta_n > -\epsilon$, so $\lim_{n\to\infty}2^{-\delta_n}$ satisfies an equivalent characterization of what it means to be the supremum of the set $\{2^r : r \in \Bbb Q\ \text{and}\ r<0\}$. The equality $\sup_{r<0} 2^r = 2^0 = 1$ is Exercise 6 from Chapter 1 of Rudin's book. Hence,
\begin{align*}
\lim_{n\to\infty}\sqrt[2n-1]{\frac{1}{2^n}} = \lim_{n\to\infty}\sqrt[2n]{\frac{1}{2^n}} \cdot \frac{1}{2^{\delta_n}} = \lim_{n\to\infty}\sqrt[2n]{\frac{1}{2^n}}\cdot \lim_{n\to\infty}\frac{1}{2^{\delta_n}} = \frac{1}{\sqrt 2}\cdot 1 = \frac{1}{\sqrt 2},
\end{align*}
as desired.
A: Yes we can use it writing
$$\sqrt[2n-1]{\dfrac{1}{2^n}}
=\frac{1}{2^{\frac{n-\frac12}{2n-1}}\cdot2^{\frac{\frac12}{2n-1}}}
= \sqrt[2n-1]{\dfrac{1}{2^{n-\frac12}}}\cdot \sqrt[2n-1]{\dfrac{1}{\sqrt2}}$$
but it doesn't seem to be the best method.
