Finding the limit of $\left(\dfrac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n$ I'm trying to solve this limit, for which I already know the solution thanks to Wolfram|Alpha to be $\sqrt[3]{abc}$:
$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n:\forall a,b,c\in\mathbb{R}^+$$
As this limit is an indeterminate form of the type $1^\infty$, I've been trying to approach it by doing:
$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}\cdot\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}=e^{\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}$$
But now when I approach that top limit this is what I get:
$$\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n=\lim_{n\rightarrow\infty}\frac{n\cdot a^{\frac{1}{n}}}{3}+\frac{n\cdot b^{\frac{1}{n}}}{3}+\frac{n\cdot c^{\frac{1}{n}}}{3}-n=\lim_{n\rightarrow\infty}\frac{n\cdot a^0}{3}+\frac{n\cdot b^0}{3}+\frac{n\cdot c^0}{3}-n=\lim_{n\rightarrow\infty}\frac{n}{3}+\frac{n}{3}+\frac{n}{3}-n=0$$
And hence the final limit should be $e^0=1$ which is clearly wrong but I honestly don't know what I did wrong, so what do you suggest me to solve this limit?
 A: By Taylor series we have: 
$$\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}=\frac 1 3\left(3+\frac1 n(\log a +\log b+\log c)++o\left(\frac 1 n\right)\right)=1+\frac 1 n \log\sqrt[3]{abc}+o\left(\frac 1 n\right)$$
so 
$$\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\exp\left(n\log\left(1+\frac 1 n \log\sqrt[3]{abc}+o\left(\frac 1 n\right)\right)\right)\sim_\infty \sqrt[3]{abc} $$
A: While $\sqrt[n]a\to 1$, it is not correct to say $(n\sqrt[n]a-n)\to 0$. Actually, $\sqrt[n]{1+\epsilon}\approx 1+\frac1n\epsilon$ so $n\sqrt[n]a-n\approx a$.
A: This is not an answer to the question itself I think, but still a useful observation.
This can be found using the Generalized means. Say that
$$
M_p(x_1,\dots,x_n)=\left(\frac 1n\sum x_i^p\right)^\frac 1p
$$
In general, we have $M_m(x_1,\dots,x_n)\geq M_n(x_1,\dots,x_n)$ when $m\geq n$, with equality if and only if $x_i=x_j$ for all $i$ and $j$. Also, (I now realize THIS is your question, although I'll just assume it's true), we have
$$M_0(x_1,\dots,x_n)=\sqrt[n]{x_1\cdots x_n}$$
Thus, we have
$$\lim_{p\to 0} M_p(a,b,c)=M_0(a,b,c)=\sqrt[3]{abc}$$
A: Fact I. 
$$
\lim_{n\to\infty}\left(1+\frac{a}{n}+\frac{b}{n^2}\right)^{\!n}=\mathrm{e}^a.
$$
Fact II. For every $a>0$, there exists a $b>0$, such that
$$
1+\frac{\ln a}{n}\le a^{1/n} \le 1+\frac{\ln a}{n}+\frac{b}{n^2}.
$$
Using the two facts:
$$
1+\frac{\ln a+\ln b+\ln c}{3n}\le\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\le
1+\frac{\ln a+\ln b+\ln c}{3n}+\frac{k}{n^2}
$$
and hence
$$
\left(1+\frac{\ln\sqrt[3]{abc}}{n}\right)^{\!n}\le
\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^{\!n}
\le\left(1+\frac{\ln(\sqrt[3]{abc})}{n}+\frac{k}{n}\right)^{\!n}
$$
and as the both the left and right hand side tend to $\sqrt[3]{abc}$, so does the middle one.
