Proving that: $\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$

You can use the following inequality:

$$ \sqrt{xy} \le \frac{x+y}{2} \le \sqrt[x+y]{x^x y^y}$$

The first inequality is straightforward, and the second one can be gotten by

$$ \frac{2}{x+y} = \frac{ x \times 1/x + y \times 1/y}{x+y} \ge \sqrt[x+y]{\frac{1}{x^x y^y}}$$

using the weighted $\text{AM} \ge \text{GM}$.

Setting $x = a^{1/n}$, $y = b^{1/n}$ and taking the $n^{th}$ powers gives us that the limit is $\sqrt{ab}$, by the squeeze theorem.

An elementary proof. We use the Taylor series $e^x = 1 + x + O(x^2)$ and the fact that $\lim_{n\to\infty}(1+x/n)^n = e^x$.

If $a=b$ the identity is trivial. Without loss of generality, assume $0<a<b$. Then $$\begin{eqnarray*} \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n &=& b \left(\frac{1+(\frac{a}{b})^{1/n}}{2}\right)^n \\ &=& b \left(\frac{1+e^{\frac{1}{n}\ln \frac{a}{b}}}{2}\right)^n \\ &=& b \left(1+\frac{1}{2}\frac{1}{n}\ln \frac{a}{b} + O(1/n^2)\right)^n \\ &=& b \left(1+\frac{1}{n}\ln \sqrt{\frac{a}{b}}\right)^n + O(1/n). \end{eqnarray*}$$ Therefore, $$\begin{eqnarray*} \lim_{n\to\infty} \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n &=& \lim_{n\to\infty} b \left(1+\frac{1}{n}\ln \sqrt{\frac{a}{b}}\right)^n \\ &=& b e^{\ln \sqrt{\frac{a}{b}}} \\ &=& \sqrt{a b}. \end{eqnarray*}$$

For large $n$, $$ x^{1/n}=1+\frac1n\log(x)+O\left(\frac{1}{n^2}\right) $$ Thus, $$ \begin{align} \lim_{n\to\infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n &=\lim_{n\to\infty}\left(1+\frac1n\left(\frac{\log(a)+\log(b)}{2}\right)+O\left(\frac{1}{n^2}\right)\right)^n\\ &=\exp\left(\frac{\log(a)+\log(b)}{2}\right)\\ &=\sqrt{ab} \end{align} $$