On a simpler note, if you take WLAN systems, the preamble has initial sequence called STF which is comprised of repetitions of known sequences. Let us say the length of basic sequence is $L$. If there are 4 repetitions (like 802.11n), total length of STF will be $5/4L$ including Cyclic Prefix. For now, we will ignore cyclic prefix part and let us consider only 2 repeating parts of total length $2L$. The exponential factor on each time domain symbol due to frequency offset is $e^{-j2\pi\Delta fn/F_s}$, where $\Delta f$ is the frequency offset in $Hz$ and $F_s$ is the symbol rate. In order to estimate coarse frequency offset, we multiply first half with conjugate of second and take average of phase of this multiplication. $w_n$ is the manifestation of phase noise + WGN at the receiver.
$$
\phi_n = phase(|x[n]|e^{j\theta_n}e^{-j2\pi\Delta fn/F_s}.*conj(|x[n+L])|e^{j-\theta_n}e^{-j2\pi\Delta f(n+L)/F_s})) \\
\phi_n = 2\pi \Delta fL/F_s + w_n\\
\hat{\phi} = \frac{1}{L}\sum_{n=0}^{n=L-1}\phi_n\\
\hat{\Delta f} = \frac{F_s\hat{\phi}}{2\pi L }
$$
Now, if the frequency offset is too high, the phase of exponential $2\pi \Delta f L/F_s$ should not exceed $\pm \pi$ at the edge of one basic sequence. If it exceeds, it will wrap around and will cause the above estimation to be incorrect. So
$$
2\pi \Delta fL/F_s \le |\pi|\\
\Delta f \le |F_s/(2L)|
$$
If you use a repeating sequence of $L$ length in the preamble, then the maximum frequency offset you can tolerate is
$$
\Delta f_{max} =|F_s/(2L)|$$.
