Easiest way to show $ \lim \limits_{x\to 0} \frac { |x_1|^{\alpha_1} \dotsb |x_n|^{\alpha_n} } {\| x\|^p} \text { exists } \iff \sum{\alpha_i} > p$ What is the easiest way to show $$ \lim \limits_{x\to0} \frac { |x_1|^{\alpha_1} \dotsb |x_n|^{\alpha_n} } {\| x\|^p} \text { exists } \iff \sum{\alpha_i} > p, \,\,\,\,\text {   for } \alpha_i >0 \text { ?} $$ 
One way would be to quote this result, which gives a similar theorem except that the denominator is $\sum |x_i|^{\beta_i}$, and take $\beta_i=p$ for all $i$, and note that the $2$ norm and the $p$ norm are equivalent, so $\|x\|_p^p \lesssim \|x\|_2^p$ and vice-versa.
Anyone know a simpler way? 
We'd like to maybe switch to spherical coordinates and claim that the trigonometric factor we're left with in the numerator is bounded above, but why is it bounded?
 A: $(\Leftarrow)$ Notice that
$$|x_i|\leq ||x||$$
so
 $$0 \leq\frac { |x_1|^{\alpha_1} \dotsb |x_n|^{\alpha_n} } {\| x\|^p}\leq ||x||^{\sum\alpha_i-p}\to0 $$
$(\Rightarrow)$
If 
$$|x_i|=t,\quad\forall i$$
then
$$\lim_{x\to0}\frac { |x_1|^{\alpha_1} \dotsb |x_n|^{\alpha_n} } {\| x\|^p}=\lim_{t\to0^+}\frac{t^{\sum_i\alpha_i}}{nt^p}=\lim_{t\to0}\frac{1}{n}t^{\sum_i\alpha_i-p}<\infty\iff \sum_i\alpha_i-p\geq0$$
and if $\sum_i\alpha_i=p$ then the above limit is $\frac{1}{n}$ and by choosing $x_1=0$ the limit becomes $0$ so if the limit exists then $\sum_i\alpha_i-p>0$ QED
A: Note that $${{{\sqrt {x_1^2 +  \cdots  + x_n^2} }}}\leq\sqrt n \max_{1\leq i\leq n}|x_i|$$
And for each $i$ $$\left| {{x_i}} \right| \leqslant \sqrt {x_1^2 +  \cdots  + x_n^2} $$
Thus, for one $$0 \leqslant \frac{{|{x_1}{|^{{\alpha _1}}} \cdots |{x_n}{|^{{\alpha _n}}}}}{{{{\left\| {\bf{x}} \right\|}^p}}} \leqslant \frac{{{{\left\| {\bf{x}} \right\|}^{\sum {{\alpha _i}} }}}}{{{{\left\| {\bf{x}} \right\|}^p}}}={\left\| {\bf{x}} \right\|^{\sum {{\alpha _i} - p} }}$$
and this proves one direction. For the other direction, your idea is fine. In fact, as Sami cleverly points out, setting $x_i=t$ works just as fine.
