limit $ \lim \limits_{n \to \infty} {\left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n} $

Assume $z>0$. One may write, as $n \to \infty$, $$ \begin{align} z^{1/\sqrt n}=e^{(\log z)/\sqrt n}&=1+\frac{\log z}{\sqrt n}+\frac{(\log z)^2}{2n}+O\left(\frac1{n^{3/2}} \right)\\ z^{-1/\sqrt n}=e^{-(\log z)/\sqrt n}&=1-\frac{\log z}{\sqrt n}+\frac{(\log z)^2}{2n}+O\left(\frac1{n^{3/2}} \right) \end{align} $$ giving $$ \frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}=1+\frac{(\log z)^2}{2n}+O\left(\frac1{n^{3/2}} \right) $$ and, as $n \to \infty$, $$ \begin{align} \left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n&=\left(1+\frac{(\log z)^2}{2n}+O\left(\frac1{n^{3/2}} \right)\right)^n\\\\ &=e^{(\log z)^2/2}+O\left(\frac1{n^{1/2}} \right) \to e^{(\log z)^2/2} \end{align} $$

Set $a = \log z$. Using the continuity of $\log$ we take the logarithm of the limit, so that we only need to calculate \begin{equation} \lim_{n\to \infty} n\log(\frac{e^{a\frac{1}{\sqrt{n}}}+e^{-a\frac{1}{\sqrt{n}}}}{2}) = \lim_{n\to \infty} n \log \cosh(\frac{a}{\sqrt{n}}). \end{equation} We have \begin{equation} \lim_{x\to 0}\frac{\log\cosh(ax)}{x^2} = \lim_{x\to 0}\frac{\frac{a\sinh(ax)}{\cosh(ax)}}{2x} = \frac{a}{2}\lim_{x\to 0}\frac{\sinh(ax)}{x} = \frac{a^2}{2}, \end{equation} Hence \begin{equation} \lim_{n\to \infty} n \log \cosh(\frac{a}{\sqrt{n}})= a^2/2. \end{equation} Taking the exponent we obtain the result.

$e^x$ = $1 + x + x^2 / 2 + x^3 / 3! + x^4 / 4! ...$

$e^{-x}$ = 1 - x + x^2 / 2 - x^3 / 3! + x^4 / 4! - + ...$

$((e^x + e^{-x}) / 2)$ = $1 + x^2 / 2 + x^4 / 4! + x^6 / 6! ...

In your case, x = $ln z / n^{1/2}$, which makes

$((e^x + e^{-x}) / 2)$ = $1 + ln^2 z / 2n + ln^4 z / 4!n^2 + ln^6 z / 6!n^3 ... $

To raise to the n-th power, we take the logarithm, about $ln^2 z / 2n + O (1/n^2)$, multiply by n giving about $ln^2 z / 2 + O (1/n)$, exponentiate giving about $exp(ln^2 z / 2 + O (1/n))$ so the limit is $exp(ln^2 z / 2)$ or $z^ {ln z / 2}$