TITLE: $\lim_{n\to \infty}\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n\stackrel{?}{=}\sqrt{ab}$
QUESTION [8 upvotes]: I found this interesting equality, but I could not find a way to prove it. Any (beautiful) idea?
$$\lim\limits_{n\to \infty}\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n=\sqrt{ab}$$

REPLY [6 votes]: 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 $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\lim_{n \to \infty}n(x^{1/n} - 1) = \log x$$