Geometric Mean limit of $\ell_p$ norm of sums My analysis professor introduced the $\ell_p$ norm to our class as:
\begin{align}
\| x \|_p = \left(\frac{1}{n}\sum_{j=1}^{n} |x_j|^p\right)^{1/p}.
\end{align}
We are asked to prove the following:
\begin{align}
\lim_{p \to 0} \|x\|_p &= \left( \prod_{j=1}^{n} |x_j| \right)^{1/n},
\end{align}
Can anyone give me a sort of "intuition" as to why this is true, and a hint as to how to approach the problem? All the reading material I come across uses measure theory and integrals instead of sums so I can't quite follow it.
On another note, is there any way to improve mathematical intuition? It seems that every proof in the class relies on little mathematical "tricks", which I find frustrating because I don't even know where to begin for the problems we're assigned. (I find that it's not at all the same in my Algebra or Probability classes)
Thank you!
 A: Consider the following:
$$
\begin{align}
f(p)=\| x \|_p = \left(\frac{1}{n}\sum_{j=1}^{n} |x_j|^p\right)^{1/p}
\end{align}
$$
It is well known that $f(p)$ for $p>0$ is increasing function of $p$. For $p<q$:
$$
f(p)=\| x \|_p = \left(\frac{1}{n}\sum_{j=1}^{n} |x_j|^p\right)^{1/p}\leq f(q)=\| x \|_q = \left(\frac{1}{n}\sum_{j=1}^{n} |x_j|^q\right)^{1/q}
$$
So be tending $p$  zero, we should expect that we get a function which is independent of $p$ and is less or equal than $f(p)$ for all $p>0$. One particular candidate can be geometric mean because using arithmetic-geometric inequality we have for all $p>0$:
$$
f(p)=\| x \|_p = \left(\frac{1}{n}\sum_{j=1}^{n} |x_j|^p\right)^{1/p}\geq \left( \prod_{j=1}^{n} |x_j|^p \right)^{1/pn}=\left( \prod_{j=1}^{n} |x_j| \right)^{1/n}. 
$$ 
To prove this limit, as it is suggested in one comment, you can calculate the limit of $\log f(p)$ as $p\to 0$ and use Hopital rule.
A: Assume that every $x_j\ne0$, when $p\to0$, $|x_j|^p\to1$ for each $j$ hence $\|x\|_p^p=\frac1n\sum\limits_{j=1}^n|x_j|^p\to1$. Assuming temporarily that 
$$
|x_j|^p=1+py_j+o(p)
$$ 
for every $j$ yields $\|x\|_p^p=1+py+o(p)$ with $y=\frac1n\sum\limits_{j=1}^ny_j$. Using the fact that $(1+py)^{1/p}\to\mathrm e^y$, one would get $\|x\|_p\to\mathrm e^y=\left(\prod\limits_{j=1}^n\mathrm e^{y_j}\right)^{1/n}$. Now, $|x_j|^p=\exp(p\log|x_j|)=1+py_j+o(p)$ with $y_j=\log|x_j|$ hence indeed the assumption above holds and $\mathrm e^y=\left(\prod\limits_{j=1}^n|x_j|\right)^{1/n}$.
If at least one $x_j$ is zero, replace each $x_j=0$ by some positive $\varepsilon$ and use the above to show that $\limsup\limits_{p\to0}\|x\|_p\left(\prod\limits_{j=1}^nu_j^\varepsilon\right)^{1/n}$ where $u_j^\varepsilon=\varepsilon$ if $x_j=0$ and $u_j^\varepsilon=|x_j|$ otherwise. Then let $\varepsilon\to0$.
