<s>[INST] Proving that: $\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$ Let $a$ and $b$ be positive reals. Show that
$$\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$$
 [/INST] Let us first consider that case that $ab=1$, i.e., $b=1/a$. So we want to show find the limit
$$\lim\limits_{n\to\infty} \left(\frac {a^{1/n}+\frac1{a^{1/n}}}2\right)^n = 1. \tag{1}$$
W.l.o.g. we may assume that $a\ge1$. (Otherwise we replace $a$ by $1/a$.)
We can use the fact that for $x\ge1$ we have $2\le x+\frac1x \le 2+(x-1)^2$, which can be verified by a straightforward algebraic manipulation.
(Just notice that $x+\frac1x-2 = \frac{x^2-2x+1}x = \frac{(x-1)^2}x$ and $2+(x-1)^2-x-\frac1x= \frac{x^3-3x^3+3x-1}x = \frac{(x-1)^3}x$.)
So we have
$$1 \le \frac{x+\frac1x}2 \le 1+\frac12(x-1)^2\tag{2}$$
for $x\ge1$.
Using $(2)$ for $x=a^{1/n}$ we get
$$1 \le \left(\frac {a^{1/n}+\frac1{a^{1/n}}}2\right)^n \le \left(1+\frac12(a^{1/n}-1)^2\right)^n \le 1+\frac n2 (a^{1/n}-1)^2.\tag{3}$$
Now it suffices to notice that
$$\lim\limits_{n\to\infty} \frac n2 \left(a^{1/n}-1\right)^2 = 
\lim\limits_{n\to\infty} \frac 1{2n} \left(\frac{a^{1/n}-1}{1/n}\right)^2 = 0.$$
(Here we are using that $\lim\limits_{x\to0} \frac{a^x-1}x = \ln a$, which implies that $(\frac{a^{1/n}-1}{1/n})^2$ is bounded.)
So by squeeze theorem we get that all expressions in $(3)$ converge to $1$ for $n\to\infty$, which proves $(1)$.

Once we have $(1)$ we can prove general case using
$$\left(\frac{a^{1/n}+b^{1/n}}2\right)^n = \sqrt{ab} \left(\frac{\left(\sqrt{\frac{a}{b}}\right)^{1/n}+\left(\sqrt{\frac{b}{a}}\right)^{1/n}}2\right)^n.$$
We get the same limit as in $(1)$ with $\sqrt{\frac{a}{b}}$ instead of $a$.
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