Find the limit $ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$ For $a,b,c>0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$
how can I find the limit of sequence above? 
Provide me a hint or full solution.
thanks ^^
 A: $$\lim_{n \to \infty}\ln \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n=\lim_{n \to \infty}\frac{\ln \left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)}{\frac{1}{n}}=\lim_{x\to 0^+}\frac{\ln \left(\frac{a^{x}+b^{x}+c^{x}}{3}\right)}{x}$$
L'Hospital or the definition of the derivative solves it. Actually since the limit is just the definition of the derivative of $\ln \left(\frac{a^{x}+b^{x}+c^{x}}{3}\right)$ at $x=0$, it would be wrong to use L'H :)
A: This answer is inspired by @Aryabhata's answer here. 
Suppose $x_k>0$. 
By $\mathrm{AM}\ge\mathrm{GM}$ we have 
$$\sqrt[m]{x_1\cdots x_m} \le \frac{x_1+\ldots +x_m}{m}.$$
By weighted $\mathrm{AM}\ge\mathrm{GM}$ we have 
$$\begin{eqnarray*}
\frac{m}{x_1+\ldots +x_m} 
&=& \frac{x_1\cdot 1/x_1+\ldots +x_m\cdot 1/x_m}{x_1+\ldots +x_m} \\
%&\ge& \left(\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}\right)^{1/(x_1+\ldots +x_m)} \\
%&\ge& \left(\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}\right)^{\frac{1}{x_1+\ldots +x_m}} \\
&\ge& \sqrt[x_1+\ldots +x_m]{\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}}.
\end{eqnarray*}$$
Therefore, 
$$\sqrt[m]{x_1\cdots x_m} 
\le \frac{x_1+\ldots +x_m}{m}
\le \sqrt[x_1+\ldots +x_m]{x_1^{x_1}\cdots x_m^{x_m}}.$$
Letting $x_k = a_k^{1/n}$ and taking the $n$th power we find by the squeeze theorem that 
$$\lim_{n \to \infty}\left(\frac{a_1^{1/n}+\ldots +a_m^{1/n}}{m}\right)^n
= \sqrt[m]{a_1\cdots a_m}.$$
In particular, 
$$\lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n
= \sqrt[3]{a b c}.$$ 
A: $$\begin{align}\lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^{n} &=\lim_{r \to 0^+}\left(\frac{a^{r}+b^{r}+c^{r}}{3}\right)^{1/r}\\ &=\lim_{r \to 0^+}\frac{1}{r}\cdot \exp\left( [\log \left(a^{r}+b^{r}+c^{r}\right)]\right)
\\
&=\exp\left( [\lim_{r \to 0}\frac{d}{dr}\log(a^{r}+b^{r}+c^{r})]\right)
\\
&=\exp\left( \frac{[\log(a)+\log(b)+\log(c)]}{3}\right) 
\\
&= (abc)^{1/3}
\end{align}$$
A: $a^{\frac 1n}=1 +\dfrac{\log a}{n}+O(\frac1{n^2})$
So,$\Big(\dfrac{a^{\frac 1 n}+b^{\frac 1 n}+c^{\frac 1 n}}3\Big)^n=(1+\frac{\log abc}{3n}+O(\frac1{n^2}))^n\to \sqrt[3]{abc}$ This generalizes easily to 4 or more terms also.
