How does Vega of a call/put behave under the Black-Scholes model?

I have two questions. I would prefer a reference if possible.

1. Is the value of vega bounded for $\sigma\in [0,\infty)$? (I assume so, I imagine it goes to 0 as $\sigma$ go to infinity.)

2. Are there any well established properties of vega? (e.g. convexity/monotonicity? How fast does it go to 0, as $\sigma\rightarrow\infty$? How does it behave as $\sigma\rightarrow 0$?)

A partial answer would be more than welcome.

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if we write down the formula for vega we get $$\text{vega} = S \sqrt{T} e^{-q T} \frac{1}{\sqrt{2\pi}} \exp(- \frac{d_1^2}2)$$ where $$d_1 = \frac{\log(S/K) + (r-q +\sigma^2/2)T}{\sigma \sqrt{T}}.$$ We have $$\lim_{\sigma \rightarrow \infty} d_1 = \infty,$$ as $\sigma^2$ grows quicker than $\sigma$ for $\sigma \rightarrow \infty$ and therefore $$\lim_{\sigma \rightarrow \infty} \text{vega} = 0,$$ due the definition above and the continuous functions involved.
Furthermore $$\lim_{\sigma \rightarrow 0} d_1 \text{ is unbounded},$$ because $$\lim_{\sigma \rightarrow 0} d_1 = \lim_{\sigma \rightarrow 0} \frac{\log(S/K)}{\sigma \sqrt{T}} + \frac{(r-q) \sqrt{T}}{\sigma} +\sigma/2 \sqrt{T}$$ and the first two summands are unbounded for $\sigma \rightarrow 0$ and the third tends to zero. Thus depending on $\log(S/K)$ - the moneyness - and $r-q$ we get $$\lim_{\sigma \rightarrow 0} d_1 = \pm \infty$$ and in any case $$\lim_{\sigma \rightarrow 0} \text{vega} = 0.$$
For the second part of convexity/monotonicity you need $\frac{\partial \text{vega}}{\partial \sigma}$ this is called Volga.
@Lost1 I mean that depending on the constellation of $\log(S/K)$ and $r-q$ which determines the sign we get either $+\infty$ or $-\infty$. But as $d_1$ is squared the sign does not matter. – Richard Jul 18 '14 at 8:10