To put it in more qualitative physical form: Of course, in a strict sense there's no perfectly exact measurement of position in the sense that you determine it to a geometrical point, i.e., you can measure the position of a particle only with a finite resolution, but there is no fundamental principle that sets a limit to this resolution. So given a particle prepared in some quantum state ##\hat{\rho}## (which may be a pure or mixed state) to measure the ##\Delta x## of the uncertainty relation you have to measure the position of the particle at a resolution much higher than ##\Delta x##, and you have to repeat this sufficiently often on an ensemble of particles prepared in the specific state ##\hat{\rho}## such that you get the standard deviation ##\Delta x## at sufficient statistical significance (what's "sufficient" is defined by the experimentalist of course).
The same holds true for a single-momentum measurement, and what's meant by the textbook Heisenberg uncertainty relation ##\Delta x \Delta p \geq 1/2## refers to such single-position and single-momentum measurements each with a resolution much better than ##\Delta x## and ##\Delta p##, respectively. It's not referring to a "simultaneous measurement of postion and momentum" nor to its theoretical possibility or impossibility.
To make statements about such a simultaneous measurement and how accurate you can make it in principle, you have to specify concrete measurement devices to do so.
One can, however, also look at it at from the point of view of information theory. Then the formalism gives you a theoretical limitation for any such simultaneous measurements. The argument is not that difficult. You suppose that you have measured somehow position and momentum on an ensemble such that you know that the average position and momentum are ##\langle x \rangle## and ##\langle p \rangle## and the standard deviations are ##\Delta x^2## and ##\Delta p^2##. From the Shannon-Jaynes principle of maximum entropy this gives a statistical operator of minimal prejudice (maximum entropy),
$$\hat{\rho}=\frac{1}{Z} \exp[-\lambda_1 (\hat{x}-x_0)^2-\lambda_2 (\hat{p}-p_0)^2 ],$$
where ##\lambda_1##, ##\lambda_2##, and ##x_0## and ##p_0## are convenient choices of Lagrange parameters to fulfill the conditions that the expectation values and standard deviations of position and momentum take the given values. You find the details of the further calculation in
https://itp.uni-frankfurt.de/~hees/publ/stat.pdf
It turns out that you have to set ##x_0=\langle x \rangle## and ##p_0=\langle p \rangle##. The partion sum is
$$Z=\frac{1}{2 \sinh(\sqrt{\lambda_1 \lambda_2}/2},$$
and the standard deviations simply
$$\Delta x^2=-2 \partial_{\lambda_1} \ln Z=\sqrt{\frac{\lambda_1}{4 \lambda_2}} \coth \left(\frac{\sqrt{\lambda_1 \lambda_2}}{2} \right),$$
$$\Delta p^2=-2 \partial_{\lambda_2} \ln Z=\sqrt{\frac{\lambda_2}{4 \lambda_1}} \coth \left(\frac{\sqrt{\lambda_1 \lambda_2}}{2} \right).$$
This implies that
$$\Delta x \Delta p = \frac{1}{2}\coth \left(\frac{\sqrt{\lambda_1 \lambda_2}}{2} \right) \geq \frac{1}{2},$$
i.e., no matter how accurately you measure simultaneously position and momentum, their standard deviations fulfill the Heisenberg uncertainty relation.
