For what values of $\beta$ does $\omega_i^2 \mid \omega \mid^{-\beta}$ not blow up at the origin?

I'm trying to show that the expression

$\omega_i^2 \mid \omega \mid^{-\beta}$

(where $\mid \omega \mid = ( \omega_1^2 + \omega_2^2)^{\frac{1}{2}}$ and $i = [1, 2]$) is defined at the origin under certain conditions. Here's my attempt. I'd like to know if its correct.

Because $\mid \omega \mid^2 = \omega_1^2 + \omega_2^2 \Rightarrow \omega_i^2 \le \mid \omega \mid^2$ for any $i$. Therefore, $\omega_i^2 \mid \omega \mid^{-\beta} \le \mid \omega \mid^2 \mid \omega \mid^{-\beta} = \mid \omega \mid^{2 - \beta}$.

Therefore the expression is defined at the origin if $2 - \beta > 0$ or $\beta < 2$.

Edit: in addition can it be shown that because $\omega_i \le \mid \omega \mid$ (I assume this is obvious), $\omega_i^3 \le \mid \omega \mid^3$ and by combining both inequalities it is possible to show in general that $\omega_i^n \le \mid \omega \mid^n$ for odd and even values of $n$.

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$\omega_i \lvert\omega\rvert^\beta$ is continuous at the origin if $\beta > -1$. You seem to have changed your expression to $\omega_i^2 \lvert\omega\rvert^\beta$ in your second paragraph, and you erroneously said $\lvert\omega\rvert^2\lvert\omega\rvert^\beta = \lvert\omega\rvert^{2-\beta}$ when the exponent should be $2+\beta$ instead. –  Rahul Nov 12 '10 at 17:03
Thanks Rahul. I've fixed the typos. The expression I'm looking to analyze is $\omega_i^2 \mid \omega \mid^{-\beta}$. –  Olumide Nov 12 '10 at 17:15

Yes, with your edit, your proof is fine. I think what you are trying to show is that the function $f : \omega \mapsto \omega_i^2\lvert\omega\rvert^{-\beta}$ has a well-defined and finite limit at the origin.
To make your proof fully rigorous, all you need to say is that $f$ is bounded between two continuous functions $g : \omega \mapsto 0$ and $h : \omega \mapsto \lvert\omega\rvert^{2-\beta}$ whose gap tends to zero at the origin, therefore $f$ must also be continuous at the origin. And that's a perfectly good way to show that a function is continuous.
You can also show that the condition is necessary, i.e. that if $\beta \ge 2$ then $f$ is not continuous at the origin, simply by taking limits approaching the origin along two different directions, for example one along $\omega_1 = 0$ and the other along $\omega_2 = 0$.
Thanks. How about the remark that $\omega_i^n \le \mid \omega \mid^n$ for all positive $n$. –  Olumide Nov 12 '10 at 22:34
@Olumide: Yes, that's true too. In fact it follows from the facts that $\omega_i \le \lvert\omega\rvert$ and that $x^n$ is an increasing function on $x \ge 0$. (Except if $\omega_i < 0$, but then the proposition is trivial.) –  Rahul Nov 13 '10 at 0:04