How to calculate the input and output capacitors of a common emitter amplifier using BJT?
When the impedance of the capacitor equals that of a load resistor, the gain of the circuit is 70.71% i.e. an attenuation of 3 dB. For the load resistor this means: -


Bias resistors connected to power rails are to be regarded as being in parallel
Base current equivalent resistance is in parallel to bias resistors


If you go lower in frequency, the attenuation is more and if you go higher in frequency then gain approaches unity. So, set a cut-off frequency where an attenuation of 3 dB is of no-consequence to the performance you wish to obtain.


  Several authors say to calculate this capacitor equal to the input
  impedance at the cut frequency.


That depends on what you are trying to achieve but I would play safe and set it at a frequency where my lowest desired signal frequency will hardly be attenuated and, in many cases, 3 dB attenuation might be regarded as too much.


  Some authors say the frequency must be the lowest frequency, below
  which, I want to remove from the signal. Some say it must be the highest frequency the amplifier must amplify.


That would be true of a low-pass filter. The filter in your question is a high pass filter.