Using AM-GM to prove $\lim_{p\to 0}$ of the $p$-mean is equal to the GM. The problem in question is.

Let $a_1,a_2,...,a_n>0$. Prove that
   $${\lim_{p\to 0} \left(\frac{{a_1}^p+{a_2}^p+\cdots+{a_n}^p}{n}\right)^{1/p}}=(a_1a_2\cdots a_n)^{1/n}$$ i.e. show that as $p$ approches $0$, the $p$-mean approaches the geometric mean. 

The direct proof of this is simple and can be done using logorithms. However when I started this problem I had the following idea: 
Notice first that by the $AM-GM$ inequality,   
$$\frac{a_1^p+\cdots+a_n^p}{n}\geq(a_1a_2\cdots a_n)^{p/n}$$ Thus, $$\left(\frac{a_1^p+\cdots+a_n^p}{n}\right)^{1/p}\geq(a_1a_2...a_n)^{1/n}$$ Equality is achieved in the above iff $a_1^p=a_2^p=\cdots=a_n^p$. But when $p$ approaches $0$, $a_1^p=a_2^p=\cdots=a_n^p=1$. Of course this does not prove the problem in question but I am trying to use this idea to construct a proof but it is much trickier that expected. Any ideas? The most promising idea I had so far was to create a subsequence of the p-mean and prove that it is decreasing and the geometric mean is the infinimum of this subsequence. But that is very hard to do. 
So I guess my main question is:
Could we find some special case where if the terms $a_1,a_2,\cdots,a_n$ approached some $x$ in some limit $L$ then, $L(AM)$=$L(GM)$?
 A: I have no proof using AM-GM alone.  But this answer may be useful.  
We can prove a more general result.  Let $a_1,a_2,\ldots,a_n>0$ and $w_1,w_2,\ldots,w_n> 0$ be such that $w_1+w_2+\ldots+w_n=1$ (in your setting $w_1=w_2=\ldots=w_n=1/n$).  Define the power mean
$$M(p)=\left(w_1a_1^p+w_2a_2^p+\ldots +w_na_n^p\right)^{1/p}$$
for all $p\ne0$.  We want to show that $\lim_{p\to 0}M(p)$ exists and equals the geometric mean $G=a_1^{w_1}a_2^{w_2}\cdots a_n^{w_n}$.  
Let $b_i=\ln a_i$.  Then $$a_i^p=\exp(pb_i)=1+pb_i+O(p^2).$$
That is
$$\sum_{i=1}^nw_ia_i^p=1+p\sum_{i=1}^nw_ib_i+O(p^2).$$
Therefore
$$M(p)=\left(\sum_{i=1}^nw_ia_i^p\right)^{1/p}=\left(1+p\sum_{i=1}^nw_ib_i+O(p^2)\right)^{1/p}.$$
Note that $\lim_{h\to 0}(1+ht)^{1/h}=e^t$.  As $p\to 0$, we get
$$M(p)\to \exp\left(\sum_{i=1}^nw_ib_i\right)=\prod_{i=1}^na_i^{w_i}=G.$$
So it makes sense to define $M(0)=G$.  With this definition, $M:\Bbb R\to \Bbb R$ is a non-decreasing function (and it is not strictly increasing iff $a_1=a_2=\ldots=a_n$).  You can also show that $\lim_{p\to\infty}M(p)=\max_{i=1}^na_i$ and $\lim_{p\to-\infty}M(p)=\min_{i=1}^na_i$.
A: Let $a_k=\exp(b_k)$. By homogeneity you may assume $\prod a_k=1$, or $\sum b_k=0$, then prove
$$ \lim_{p\to 0^+}\left(\frac{1}{n}\sum_{k=1}^{n}e^{pb_k}\right)^{1/p} = 1. $$
As long as $p\to 0^+$ we have
$$ e^{pb} = 1+ pb + o(p), $$
hence 
$$ \frac{1}{n}\sum_{k=1}^{n}e^{pb_k} = 1+\frac{p}{n}\sum_{k=1}^{n}b_k + o(p)=1+o(p) $$
and
$$ \left(\frac{1}{n}\sum_{k=1}^{n}e^{pb_k}\right)^{1/p} = \exp\left[\frac{\log(1+o(p))}{p}\right]=\exp\left[\frac{o(p)}{p}\right]=\exp\left[o(1)\right] $$
converges to $e^0=1$.
