Hi,
as already mentioned: it depends on your circuit.
Also on how you define "can I use it".
There is no black and white.
Theoretically you may switch huge capacitance, but then you have to expect huge current ... and huge power dissipation. So from the physics there is no limit.
The current is: dV/dt * C
and the power dissipaton is: 0.5 * C * V * V * f
as example:
If your applicaton uses very hard (high dV/dt) switching, the current becomes high, too. Short, but high peaks. Now if your PCB layout is not optimal and there is high stray inductance you get lots of ringing, which may cause lots of EMI and maybe makes the circuit inoperable.
The softer you switch, the more capacitance it can handle.
Idependent of dV/dt there will be power dissipation (mainly in the switching device) just to charge/discharge the capacitor. Since V is quared (in the above formula) it has the highest impact on power dissipation. frequency is given. This the higher the capacitance, the higher the power dissipation.
So if your application is used for high power then maybe your switching devices have 100W of power dissipation. Then an additional of 2W because of the capacitance won´t harm. But if you have a tiny application or 100mW of load power then most certainly you daon´t like to have 3W dissipated just because of the capacitance.
In the end it´s the designer who has to decide whether the "can I use" limit is at
* 1mA, 1A or 25A..
* 10mW, 0.5W or 10W.
As long as you don´t give full application details ... I guess no one can tell you wheter this TVS is useful or not ... one even can not give a recommendation.
Klaus