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

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

I now the answer, it is $ \displaystyle e^\frac{\log^2z}{2} $, but I don't know how to prove it. It seems like this notable limit $\displaystyle \lim \limits_{x \to \infty} {\left(1 + \frac{c}{x}\right)^x} = e^c$ should be useful here. For example I tried this way: $$ (z^{1/\sqrt n} + z^{-1/\sqrt n}) = (z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)})^2 + 2 $$

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

where $ (z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)})^2 $ seems close to $ \frac{\log^2 z}{n} $.

Also we can say that $$ \left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n = e^{n \log {\left(1 + \frac{\left(z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)}\right)^2}{2}\right)}}$$ and $ \log {\left(1 + \frac{(z^{1/(2 \sqrt n)} - z^{-1/(2 \sqrt n)})^2}{2}\right)} $ can be expand in the Taylor series. But I can't finish this ways.

Thanks for the help!

Answer

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}
$$