Airy function and modified Bessel function I have got a question concerning the Airy functions in relation to the Bessel function.
From Wiki, it is possible to see how
$$ Ai(x)=\frac{1}{\pi}\sqrt{\frac{x}{3}}K_{1/3}\left(\frac{2}{3}x^{\frac{3}{2}}\right) $$
The question is: how can the Airy function retrieve a 0.3550 value when evaluated in $ x = 0 $, if
$$ K_{1/3}\left(\frac{2}{3}0^{\frac{3}{2}}\right) = \infty $$
It's probably a naive question but I would say that 
$$ Ai(0) = 0*\infty = NaN$$
looking at the above equivalence.
I thank you in advance for supporting.
 A: For fixed $\nu>0\;$ and from the reference DLMF you have the equivalence for $z$ near $0$ :
$$\operatorname{K}_{\nu}(z)\sim \frac{\Gamma(\nu)}2\left(\frac z2\right)^{-\nu}$$
From this you may deduce that near $0$ the Airy function $\operatorname{Ai}(x)$ is equivalent to :
\begin{align}
\frac{1}{\pi}\sqrt{\frac{x}{3}}\operatorname{K}_{1/3}\left(\frac{2}{3}x^{\frac{3}{2}}\right)&\sim \frac{\Gamma(1/3)}{2\pi}\sqrt{\frac{x}{3}}\left(\frac{1}{3}x^{\frac{3}{2}}\right)^{-1/3}\\
&\\
&\sim \frac{\sqrt[3]{3}\ \Gamma(1/3)}{\sqrt{3}\;2\,\pi}\\
&\\
&\sim 0.355028053887817239260063186\cdots
\end{align}
Getting even a closed form for the limit.
A: You can formulate your problem in the following way: 
compute the limit of of $\frac{f(x)}{g(x)}$ as $x\to0$ where 
$$f(x)=\sqrt x\qquad \text{and}\qquad g(x)=\frac{1}{K_{\frac13}(x^{\frac32})}.$$
To study this, you can for example use a Taylor expansion at $x=0$ for $f$ and $g$.
A: Computationally, $0 * \infty$ is $NaN$.  But we're talking math here, not computers and floating point arithmetic.  (Or, at least, I assume that's what we're talking.)  You need to take the limit,
$\displaystyle \lim_{x\rightarrow 0} \text{Ai}(x)$
