What would the RGB color value of an infinitely hot blackbody be? In other words, what is the limit of the rgb values of color temperature as temperature approaches infinity? Put differently, what is the terminal point of the Planckian locus? Is there an exact value?
There is definitely an answer, since it the point (infinite color temperature) is labeled on chromaticity diagrams- I am wondering how the rgb value of the point is found.
 A: If you make the temperature very, very high (say $>10^{5}$ K) then the visible part of the spectrum lies in the Rayleigh-Jeans tail of the Planck spectrum.
Thus:
$$B_{\nu} \simeq \frac{2\nu^2 kT}{c^2},$$
and the approximation becomes better and better as $kT \gg h\nu$. The equivalent expresson per unit wavelength interval is
$$B_{\lambda} \simeq \frac{2c kT}{\lambda^4}$$
The general problem of converting from a spectrum to RGB values is discussed here. This involves doing an integration of the spectrum, weighted by visual perception sensitivity and then converting the resulting sums into RGB values. The process is described in some detail here.
No example is given for a very hot blackbody, though some tools are provided (C programs). However, I find this site has already done the calculations for blackbodies up to 30,000K (which is probably close to an asymptotic limit and gets RGB=#9fbfff (159,191,255).
Here is a plot of RGB values versus blackbody temperature from Tanner Helland. that seems to agree closely with this result (maybe 152,185,255 at 40,000 K) and where you can see the asymptotic behaviour.

Further edit:  Wolfram Alpha has a calculator that goes up to 90,000K. This yields an RGB of 153.7,176.7,255, but given that the RGB for 30,000K is identical, I'm not sure I trust these exact values. In any case, the image below is what this looks like (courtesy of Emilio Pisanty).

A: By using the spreadsheet at http://www.brucelindbloom.com/index.html?Calc.html I am getting Apple RGB values of (110,150,242), which on my screen is a purplish blue.
A: Take two wavelengths $\lambda_1 < \lambda_2$ and use the Planck's law for the spectral radiance.
$$u(\lambda, T) = \frac{2hc^2}{\lambda^5}\frac{1}{e^\frac{hc}{\lambda k T} - 1}$$  
Then lets take the fraction of the two intensities 
$$\frac{u(\lambda_2, T)}{u(\lambda_1, T)} = 
\frac{\lambda_1^5}{\lambda_2^5} \frac{e^\frac{hc}{\lambda_2 k T} - 1}{e^\frac{hc}{\lambda_1 k T} - 1}$$ 
For infinitely hot blackbody you have to take a limit. 
$$\lim_{T \to \infty}\frac{u(\lambda_2, T)}{u(\lambda_1, T)} = \frac{\lambda_1^6}{\lambda_2^6}$$
(See here.)
So, the spectrum will follow the 6-power law. That, among other things, means that for every amount of energy that $\lambda_1=700 nm$ light releases on your retina, the $\lambda_2=7 nm$ X-ray will release $10^{12}$ times more. So, converting that to RGB -- I dont think that makes much sense, since your eyes are going to be burnt.   
A: The higher the temperature, the more the peak wavelength of the radiation shifts towards higher frequencies. At higher and higher temperatures, the peak will be blue, then ultraviolet, before it shifts into X-rays and finally gamma rays.
From this you would expect that at infinite temperature, the frequency would also be infinite. In reality of course, long before it gets that high, the energy of the emitted quanta will disrupt space, as well as the black body, hence limiting itself. As a guess, when the wavelength becomes equal to the Planck length ($10^{-35}m$) the photon may collapse into a black hole. This happens when the energy reaches about $10^{19}GeV$. The frequency at that point is $3*10^{43}Hz$.
Either way, the radiation will no longer be describable in terms of RGB, as the frequencies involved will be vastly higher than those of visible light.
