$\lim_{n\to \infty}\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n\stackrel{?}{=}\sqrt{ab}$ I found this interesting equality, but I could not find a way to prove it. Any (beautiful) idea?
$$\lim\limits_{n\to \infty}\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n=\sqrt{ab}$$
 A: Let 
$$y=\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n$$
$$\implies \log y= n\log \frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}$$
Let $n=\frac{1}{p}$ as $ n \rightarrow \infty, p\rightarrow 0$
$$\implies \log y= \lim_{p \to 0}\frac{\log \frac{a^p+b^p}{2}}{p}$$
Applying L.H rule, we get
$$\log y=\lim_{p \to 0} \frac{a^p\log a+b^p\log b}{a^p+b^p}$$
$$\implies \log y=\frac {1}{2}\log ab$$
$$\implies y=\sqrt {ab}$$
A: As usual when you see the variable in exponent, the strategy is to take logs. Thus if $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{a^{1/n} + b^{1/n}}{2}\right)^{n}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(\frac{a^{1/n} + b^{1/n}}{2}\right)^{n}\text{ (via continuity of log)}\notag\\
&= \lim_{n \to \infty}n\log\left(\frac{a^{1/n} + b^{1/n}}{2}\right)\notag\\
&= \lim_{n \to \infty}n\cdot\dfrac{\log\left(1 + \dfrac{a^{1/n} + b^{1/n} - 2}{2}\right)}{\dfrac{a^{1/n} + b^{1/n} - 2}{2}}\cdot\dfrac{a^{1/n} + b^{1/n} - 2}{2}\notag\\
&= \frac{1}{2}\lim_{n \to \infty}n(a^{1/n} - 1) + n(b^{1/n} - 1)\notag\\
&= \frac{1}{2}(\log a + \log b)\notag\\
&= \frac{1}{2}\log ab\notag
\end{align}
Hence $L = \sqrt{ab}$. Here I have used two fundamental limits $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\lim_{n \to \infty}n(x^{1/n} - 1) = \log x$$
A: It is worthwhile to observe that this question is the special case of showing that the limit of the $p^{\rm th}$ power mean as $p \to 0^+$ equals the geometric mean, for two numbers $a, b > 0$.  Define the $p^{\rm th}$ power mean of a sequence of positive numbers $\boldsymbol x = (x_1, \ldots, x_m)$ for a nonzero real $p$ to be $$M_p(\boldsymbol x) = \left(\frac{1}{m} \sum_{i=1}^m x_i^p \right)^{1/p}.$$  Then we have $$\lim_{p \to 0^+} M_p(\boldsymbol x) = M_0(x) = \left(\prod_{i=1}^m x_i \right)^{1/m},$$ the geometric mean.  The proof is given in this Wikipedia article.
A: HINT:
Use 
$$\begin{align}
\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n&=\left(\frac{e^{\frac{1}{2n}(\log(a)+\log(b))}\left(e^{\frac1{2n}(\log(a)-\log(b))}+e^{-\frac1{2n}(\log(a)-\log(b))}\right)}{2}\right)^n\\\\
&=\sqrt{ab}\cosh^n\left(\frac{\log\left(\sqrt{a/b}\right)}{n}\right) \tag 1
\end{align}$$
and the squeeze theorem.
SPOILER ALERT:  Scroll over the highlighted area to reveal the solution

Starting from $(1)$, we need to evaluate the limit $$\lim_{n\to \infty}\cosh^n\left(\frac{\log\left(\sqrt{a/b}\right)}{n}\right)$$We can use the inequalities $$1\le \cosh(x)\le \frac{1}{1-x^2}$$for $|x|<1$.  Then, for $n$ sufficiently large$$1\le \cosh^n\left(\frac{\log\left(\sqrt{a/b}\right)}{n}\right)\le \left(1-\frac{\log^2\left(\sqrt{a/b}\right)}{n^2}\right)^{-n}\le \frac{1}{1-\frac1n \log^2\left(\sqrt{a/b}\right)}$$whereupon applying the squeeze theorem reveals the limit $$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n=\sqrt{ab}}$$as was to be shown!  Alternatively, Recalling that $\cosh (x)=1+O(x^2)$, then $$\begin{align}\cosh^n\left(\frac{\log\left(\sqrt{a/b}\right)}{n}\right)&=\left(1+O\left(\frac{1}{n^2}\right)\right)^n\\\\&=e^{n\log\left(1+O\left(\frac{1}{n^2}\right)\right)}\\\\&=e^{O(1/n)}\\\\&\to 1\,\,\text{as}\,\,n\to \infty\end{align}$$Finally, we have $$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n=\sqrt{ab}}$$as was to be shown!  

A: Even though there are other solutions, I just wanted to give one that is pretty simple. If we could show that this sequence is decreasing (I suspect the devil is in the details on this one) we could just employ the Arithmetic-Geometric mean inequality. We'd find:
$$ \lim_{n \to \infty} \Big(\frac{ a^{1/n} + b^{1/n} }{2}\big)^{n} \geq \lim_{n \to \infty}\sqrt{ab} = \sqrt{ab} $$
Since this is a decreasing sequence bounded below, the limit must the infimum.
