Evaluate $\lim_{n\to \infty}\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n$

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 $$\log y=\lim_{p \to 0} \frac{a^p\log a+b^p\log b}{a^p+b^p}$$ $$\implies \log y=\frac {1}{2}\log ab$$ $$\implies y=\sqrt {ab}$$

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 $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\lim_{n \to \infty}n(x^{1/n} - 1) = \log x$$

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.