What does a large capacitor do in small signal analysis?
Small signal analysis is the art of figuring out what the circuit would do if all of the DC sources went away (active elements are left at their operating points), and a small AC signal is applied at the circuit input. The (complex) impedance of a capacitor is $$Z = 1/(j\omega{}C) = 1/(j2\pi{}fC)$$

The omega is the angular frequency of your signal. For very low frequencies, this becomes a very large number, so it will act as an open [you can remove the capacitor]. 

You can also think about the high frequency limit. For high frequency signals, the impedance becomes a very small number, so it will act as a short. The exact frequency where it can be modeled as a short will depend on the rest of the circuit, but will be at a lower frequency for larger value capacitors.

But, you most likely will be interested in a frequency somewhere in the middle, so somewhere between an open and a short. In this case you would leave the capacitor in the circuit, and perhaps use the above equation for its complex impedance. For your example with parallel elements, use the "normal" parallel element equation 1/(1/Z1 +1/Z2), and use complex arithmetic. For certain frequencies, you'll see that the result is purely real (the capacitor goes away), but at other frequencies it is purely imaginary (the resistor goes away).