Limit Sandwich principle

Question

How can I find: $$\lim_{n \rightarrow \infty}\Big (\frac{a^\frac{1}{n} + b^\frac{1}{n}+ c^\frac{1}{n}}{3} \Big)^n $$ if $a > b >c >0$ and $a,b,c \in \mathbb{R}$.

I am trying to use sandwich principle. So I obtain this inequalities $$\Big( \frac{a^\frac{1}{n}}{3}\Big)^n \le \Big (\frac{a^\frac{1}{n} + b^\frac{1}{n}+ c^\frac{1}{n}}{3} \Big)^n \le \Big( \frac{3a^\frac{1}{n}}{3} \Big)^n.$$

But, I still don't get the answer to solve this problem with sandwich principle.

What should I do? Any idea? Thanks in advance.

Answer 1

$a=e^u$, $b=e^v$, $c=e^w$

$$\left(\frac{a^\frac{1}{n} + b^\frac{1}{n}+ c^\frac{1}{n}}{3} \right)^n=\left(\frac{e^\frac{u}{n} + e^\frac{v}{n}+ e^\frac{w}{n}}{3} \right)^n=$$ $$\left(\frac{1+\frac{u}{n} + 1+\frac{v}{n}+ 1+\frac{w}{n}+O\Big(\frac{1}{n^2}\Big)}{3} \right)^n=\left(1+\frac{u+v+w}{3n}+O\Big(\frac{1}{n^2}\Big)\right)^n=$$ $$\exp \frac{u+v+w}{3}+O\Big(\frac{1}{n}\Big)=\sqrt[3]{abc}+O\Big(\frac{1}{n}\Big)$$

Answer 2

$$L=\lim_{n\to \infty} \Big (\frac{a^\frac{1}{n} + b^\frac{1}{n}+ c^\frac{1}{n}}{3}\Big)^n$$ This is $1^{\infty}$ form, use $$\lim_{x \to t} f(x)^{g(x)}= \exp[\lim_{x\to t }g(x)(f(x)-1)]$$ So $$L=\exp\left[\lim_{n \to \infty}n\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}-1\right)\right]$$ $$\implies L=\exp\left[\lim_{n \to \infty}\frac{1}{3}\left(\frac{a^{1/n}-1+b^{1/n}-1+c^{1/n}-1}{1/n}\right)\right]$$ $$\implies L=\exp\left[\lim_{t \to 0}\frac{1}{3}\left(\frac{a^{t}-1+b^{t}-1+c^{t}-1}{t}\right)\right]$$ Use $\lim_{t\to 0} \frac{a^t-1}{t}=\ln a$, we get $$L=\exp[\frac{1}{3} \ln (abc)]=(abc)^{1/3}.$$