In trying to solve a certain limit, I wondered how Mathematica comes up with this weird expression for a series expansion of Jacobi's $\theta_3(0,q)$ in $q=1$ at the order 0:
$\frac{i \sqrt{\pi } e^{-i \pi \left\lfloor \frac{3}{4}-\frac{\arg (q-1)}{2 \pi }\right\rfloor }}{\sqrt{q-1}}$
As a bonus, I would like to know an intuitive (physicist-friendly) reason for which $x \vartheta _3(0,e^{-x^2})$ has a long initial plateau:
As a hint, see the section Transformations of Lattice Parameter in the NIST handbook / NIST Digital Library or the section Methods of Computation, where it is shown that the $\theta(0,q)$ series can be restricted to rather small $q.\;$ Maybe the Mathematica expansion is based on these transformations. At least I can explain your plateau observervations: For $\theta_3(q) = \theta_3(0,q)$ this transformation becomes:
$$\theta_3(q) = (-\pi/\ln q)^{1/2} \, \theta_3\big(\exp(\pi^2/\ln q)\big)$$ and with your $q = \exp(-x^2)$ or $\ln(q) = -x^2$ you get $$x\theta_3(\exp(-x^2)) = x\sqrt{\frac{\pi}{x^2}} \, \theta_3\left(\exp\left(-\frac{\pi^2}{x^2}\right)\right) = \sqrt{\pi} \, \theta_3\left(\exp\left(-\frac{\pi^2}{x^2}\right)\right).$$ Now for the range $0 < x < 0.8$ the argument for the last $\theta_3$ is very small $$ 0 < \exp\left(-\frac{\pi^2}{x^2}\right) < 0.0000002007396832$$ making the value of $\theta_3$ nearly constant, which is your long initial plateau; and because $\theta_3(0) = 1$, your plateau has the value $\sqrt{\pi} \approx 1.772454.$