Is there a way to proof that you don't see the cut-off frequency in the plot?
I don`t know exactly what your problem is. However, perhaps the following clarifies something:

With your previous question you gave the transfer function which, however, was not yet shown inthe so called "normal form", which means that the denominator of the transfer function should be D(jw)=1+jw*b+(jw)²c.....

In your case, we simply get by dividing the whole function with (R1+R2) :

H(jw)=N(jw)/D(jw) with

N(jw)=[R2/(R1+R2)][1+jwL/R2) and 

D(jw)=1+jwL/(R1+R2).

From these expressions you immediately can derive that there is one zero at wo=R2/Land a pole at wp=(R1+R2)/L.

This pole is not really identical to the 3dB cut-off but very close to the value as given elsewhere in an answer to this thread (fp=175kHz).