In a low frequency region of the single stage BJT amplifier, it is the RC combinations formed by the capacitors \$ C_{in},C_E,C_{out} \$ -Electronic Devices and Circuit Theory-Boylestad
Considering the equivalent circuit formed as shown in fig.1
The output voltage and input voltage are related by:
$$\Bbb V_{out}=\frac{\Bbb R}{\Bbb R-j\Bbb X_C}\Bbb V_{in}$$ The magnitude is given by $$V_{out}=\frac{R}{\left[ R^2+X^2_C \right]^{1/2}}V_{in}$$ $$\text{When } X_C=R$$ $$V_{out}=0.707V_{in}$$ The frequency at which this occur, is given by the equation, $$R=X_C=\frac{1}{2\pi f_LC}$$ or, 3dB cutoff frequency $$f_L=\frac{1}{2\pi RC}$$
In case the equivalent circuit formed by the BJT circuit is something like fig 2, which is the case when we are considering the input portion of the BJT circuit, the analysis is something like this:
$$\Bbb V_{in}=\frac{\Bbb R_{in}}{\Bbb R_{in}+\Bbb R_S-j\Bbb X_C}\Bbb V_{in}$$ The magnitude is given by $$V_{in}=\frac{R_{in}}{\left[ (R_{in}+R_S)^2+X^2_C \right]^{1/2}}V_{in}$$ $$\text{When } X_C=R_S+R_{in}$$ $$=>\frac{1}{2\pi f C_{in}}=R_{in}+R_S$$ $$=>f=\frac{1}{2\pi(R_{in}+R_S)C_{in}}$$
But for \$ X_C=R_S+R_{in} \$ $$V_{in}=0.707 \frac{R_{in}}{(R_S+R_{in})}V_{S}$$
Which is not $$V_{in}=0.707V_S$$
How is then the cutoff frequency equal to \$ \frac{1}{2\pi (R_S+R_{in})C_{in}}\$ in case of Figure 2 ?
Furthur,
For
$$V_{in}=0.707V_S$$
$$\frac{R_{in}}{\left[ (R_{in}+R_S)^2+X^2_C \right]^{1/2}}V_{in}=\frac{1}{2^{1/2}}$$ $$=>\frac{R_{in}}{(R_{in}+R_S)^2+X_C^2}=\frac{1}{2}$$ $$=>2R_{in}^2=(R_{in}+R_S)^2+X_C^2$$ $$=>X_C^2=(R_{in}-R_S)^2$$ $$=>X_c=R_{in}-R_S$$