Limit of $n^{\frac{1}{n}}$.

After you have shown that $L=\lim_{n\to \infty}n^{1/n}$ exists, and is positive (because every $n^{1/n}\ge 1$), then $$0<L=\lim_{n\to\infty}(2n)^{1/2n}=\lim_{n\to\infty}2^{1/2n}(n^{1/n})^{1/2}=L^{1/2}$$ because $2^{1/2n}\to 1.$

Easiest proof:

By Bernoulli,

$(1+n^{-1/2})^n \ge 1+n^{1/2} \gt n^{1/2} $.

Raising to $2/n$ power,

$n^{1/n} \lt (1+n^{-1/2})^2 =1+2n^{-1/2}+n^{-1} \lt 1+3n^{-1/2} \to 1 $.

$n^{1/n}=e^{\ln n^{1/n}}=e^{\ln n/n}$. Now exp is continuous, and $\ln n/n\to0$. So the limit is $e^0=1$.