How to determine current limiting resistor values for push pull amplifier driven by op amp?
The current from an OPA454 is just about limited to a little over 100 mA and on a good day you might get 150 mA so, R3 is chosen to prevent the OPA454 from producing this current i.e. the BJTs take over at a certain point and produce the bulk of the output current.

For instance, at 100mA drive through R3 (20 ohms) the Vbe across one of the transistors is aiming be 2V. Clearly it can't so this means the transistors are doing their job of relieving the op-amp of having to perform too hard. 

Basically, if R3 is 20 ohms and the transistors start to turn-on at (say) 1V, the peak current taken from the OPA454 is 50mA and well within the spec of the device. Remember the OPA454 cannot produce watts of power - you can't expect it to deliver 50V from a 100V rail driving 100mA - it will burn long before then.



EDITED BELOW

Here is a typical push-pull output stage that uses two more BJTs for inhibiting the drive current supplied by the output transistors: -



Taken from AllAboutCircuits useful site on push-pull amplifiers. Because the original circuit uses both bases connected together, to make this work requires maybe a 10 ohm resistor inserted in each base - basically replacing each diode in the above diagram. The pull-up and pull-down resistors are not needed.

If an emitter resistor is 0.5 ohms, at round about 2A current there will be enough voltage across it to turn on the extra transistor and this diverts base drive away from the main transistor.

Here's what it looks like in a simpler form (a clsss A single BJT driver): -



Taken from the Analog Devices Wiki website. You might also have seen this method used in controlling the current thru an LED: -