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!
 A: 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 $n \to \infty$,
$$
\begin{align}
\left(\frac{z^{1/\sqrt n} + z^{-1/\sqrt n}}{2}\right)^n&=\left(1+\frac{(\log z)^2}{2n}+O\left(\frac1{n^{3/2}} \right)\right)^n\\\\
&=e^{(\log z)^2/2}+O\left(\frac1{n^{1/2}} \right) \to e^{(\log z)^2/2}
\end{align}
$$
A: 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.
A: $e^x$ = $1 + x + x^2 / 2 + x^3 / 3! + x^4 / 4! ...$
$e^{-x}$ = 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 $z^ {ln z / 2}$
A: HINT:
If one wants to uses asymptotic analysis, then simply note that
$$\frac{z^{1/\sqrt n}+z^{-1/\sqrt n}}{2}=\cosh\left(\frac{\log(z)}{\sqrt n}\right)=1+\frac12 \frac{\log^2(z)}{n}+O(1/n^2)$$

Alternatively, we can proceed using only elementary inequalities, established without calculus, and the squeeze theorem.  To proceed, we write 
$$\begin{align}
\cosh^n\left(\frac{\log(z)}{\sqrt n}\right)&=e^{n\log\left(\cosh\left(\frac{\log(z)}{\sqrt n}\right)\right)}\\\\&=e^{n\log\left(1+2\sinh^2\left(\frac{\log(z)}{2\sqrt n}\right)\right)} \tag 1\\\\
\end{align}$$
Now, in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality that for $|x|<1$
$$1+x\le e^{x}\le \frac{1}{1-x} \tag 2$$
and 
$$\frac{x}{1+x}\le\log(1+x)\le x \tag 3$$
From $(1)$ we can easily show that
$$\frac{x}{1-x^2}\le \sinh(x)\le x \tag 4$$
for $0\le x <1$.  
Applying $(3)$ and $(4)$ to $(1)$ reveals
$$\frac{\frac12 \log^2(z)}{\left(1-\left(\frac{\log(z)}{2\sqrt n}\right)^2\right)^2} \left(\frac{1}{\cosh\left(\frac{\log(z)}{\sqrt n}\right)}\right)\le n\log\left(1+2\sinh^2\left(\frac{\log(z)}{2\sqrt n}\right)\right)\le \frac12 \log^2(z)$$
whereupon applying the squeeze theorem reveals
$$\lim_{n\to \infty}n\log\left(1+2\sinh^2\left(\frac{\log(z)}{2\sqrt n}\right)\right)=\frac12 \log^2(z)$$
Finally, exploiting the continuity of the exponential function yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{z^{1/\sqrt n}+z^{-1/\sqrt n}}{2}=e^{\frac12 \log^2(z)}}$$
A: If $L$ is the desired limit then we have
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}}}{2}\right)^{n}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(\frac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}}}{2}\right)^{n}\text{ (via continuity of log)}\notag\\
&= \lim_{n \to \infty}n\log\left(\frac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}}}{2}\right)\notag\\
&= \lim_{n \to \infty}n\cdot\dfrac{\log\left(1 + \dfrac{z^{1/\sqrt{n}} + z^{-1/\sqrt{n}} - 2}{2}\right)}{\dfrac{z^{1/\sqrt{n}}+z^{-1/\sqrt{n}} - 2}{2}}\cdot\dfrac{z^{1/\sqrt{n}}+z^{-1/\sqrt{n}} - 2}{2}\notag\\
&= \lim_{n \to \infty}n\cdot\dfrac{z^{1/\sqrt{n}}+z^{-1/\sqrt{n}} - 2}{2}\notag\\
&=\frac{1}{2}\lim_{n \to \infty}n\left(\frac{z^{1/\sqrt{n}} - 1}{z^{1/2\sqrt{n}}}\right)^{2}\notag\\
&= \frac{1}{2}\lim_{n \to \infty}\{\sqrt{n}(z^{1/\sqrt{n}} - 1)\}^{2}\notag\\
&= \frac{(\log z)^{2}}{2}\notag
\end{align}
Hence $L = \exp\left\{\dfrac{(\log z)^{2}}{2}\right\}$.
