Omitting ΔVbe in emitter follower KVL equations
As an approximation to the base emitter junction you can use a forward biased diode to predict what ΔVbe might be. For instance, a 1N4148 diode has the following characteristic: -




If you have a quiescent base current of around 100 uA DC there would be approximately 500 mV DC dropped across the diode. 
If your input signal caused that base current to alter +/- 50 uA you can see that ΔVbe would be around +/- 30 mV.
This gives rise (via a \$\beta\$ value of say 100) to an emitter current change of +/- 5 mA on a standing emitter current of 10 mA. 
If your emitter resistor is (say) 1000 ohms, that's a DC output voltage of 10 volts with a superimposed AC voltage (due to your input signal) of 10 volt p-p.


So the ΔVbe of 60 mVp-p, when compared to 10 Vp-p is 0.6% and not that big of a deal to ignore.