That formula is not in any way fundamental, the singularity at $R_i\to 0$ doesn't have much of a physical meaning, rather it means that the assumptions with which this formula was derived are violated. Namely that the voltage is nonzero.

The total resistance must be such that
$$
  U = R_{\mathrm{tot}}\cdot I_{\mathrm{tot}}.
$$
Since it's a parallel circuit, this is also
$$
  U = R_1\cdot I_1 = R_2\cdot I_2.
$$
But if one of the resistances is zero, this expression must be zero, and therefore (the total current being nonzero) we have $R_{\mathrm{tot}}=0$. Which is quite intuitive as said in Chris Gerig's comment: if there's a perfect shortcut, why would any electrons bother to take the path with nonzero resistance? They don't, so it wouldn't change anything to simply take the resistor away, in which case only the superconductor would be left, still with a resistance $0$.

Also note that a superconductor doesn't actually have zero resistance, just very much less than ordinary conductors. So you could still use the formula
$$
  \frac1{\tfrac1{R_1}+\tfrac1{R_2}}
$$
where, since $R_1\gg R_2$,
$$
  \tfrac1{R_1}+\tfrac1{R_2}\approx \tfrac1{R_2}
$$
and therefore
$$
  \frac1{\tfrac1{R_1}+\tfrac1{R_2}} \approx \frac1{\frac1{R_2}} = R_2 \approx 0.
$$

Resistance of superconducting wire in parrallel with standard wire
I have used both modes on my Canon 550D and found the results to be quite different, particularly indoors under fluorescent light. Auto white balance seems to produce yellowish images at times, while using the custom white balance seems to produce more neutral colours.

How do these two modes work, and why doesn't the auto white balance do an equivalent job?

I use an image of white objects taken with auto white balance settings in the same lighting conditions to set the custom white balance

P.S. I will try to add images to show the difference
