Wrong Output for Limit

Looking at the definition of EllipticTheta from the documentation:

$$\vartheta _3(u,q)=1+2 \sum _n^{\infty } q^{n^2} \cos (2 n u)$$

it is clear that $\vartheta_3(0, 1)$ diverges. Your problematic term is:

term = EllipticTheta[3, -((d0 g π t)/(2 l)), E^(-((d0 π^2 t)/l^2))] 

EllipticTheta[3, -((d0 g π t)/(2 l)), E^(-((d0 π^2 t)/l^2))]

Let $l = 1/s$ and replace the arguments with the small $s$ approximations:

term /. l->1/Series[s, {s, 0, 1}] //TeXForm 

$\vartheta _3\left(-\frac{1}{2} s (\pi \operatorname{d0} g t)+O\left(s^2\right),1-\pi ^2 \operatorname{d0} s^2 t+O\left(s^3\right)\right)$

which is of the form:

EllipticTheta[3, -b s, 1 - a s^2] 

When $b=0$, Mathematica knows about the asymptotic behavior of this function as $s \to 0$:

Assuming[a>0, Series[EllipticTheta[3, 0, 1 - a s^2], {s, 0, 0}]] //TeXForm 

$\frac{\sqrt{\pi } (-1)^{\left\lfloor -\frac{\arg (s)}{\pi }\right\rfloor } e^{-i \pi \left\lfloor \frac{3}{4}-\frac{\arg \left(-a s^2\right)}{2 \pi }\right\rfloor }}{\sqrt{a} s}$

However, when the second argument also depends on $s$, there appears to be a bug:

Series[EllipticTheta[3, -b s, 1 - a s^2], {s, 0, 0}, Assumptions->a>0] //TeXForm 

$\vartheta _3(0,1)+O\left(s^1\right)$

There is an internal function that is designed to handle EllipticTheta series, but it doesn't seem to be used. We can work around the above issue by using it directly. First, let's reproduce the known result above:

Assuming[a>0, System`SeriesDump`ThetaFunction[3, 0, 1-a s^2, s, 0, 0]] //TeXForm 

$\frac{\sqrt{\pi } (-1)^{\left\lfloor -\frac{\arg (s)}{\pi }\right\rfloor } e^{-i \pi \left\lfloor \frac{3}{4}-\frac{\arg \left(-a s^2\right)}{2 \pi }\right\rfloor }}{\sqrt{a} s}$

Now, let's try it on the second example:

series = Assuming[ a>0 && b>0, System`SeriesDump`ThetaFunction[3, -b s, 1-a s^2, s, 0, 0] ]; series //TeXForm 

$\frac{\sqrt{\pi } (-1)^{\left\lfloor -\frac{\arg (s)}{\pi }\right\rfloor } \exp \left(b^2 s^2 \left(\frac{1}{2}-\frac{1}{a s^2}\right)-i \pi \left\lfloor \frac{3}{4}-\frac{\arg \left(-a s^2\right)}{2 \pi }\right\rfloor \right)}{\sqrt{a} s}$

Much better! Now, recall that the above EllipticTheta term comes with a prefactor of $\sqrt{\frac{1}{l^2}}$, or equivalently $\sqrt{s^2}$, so we want the limit of s series as $s \to 0$. Hence:

lim[a_, b_] = Assuming[ a>0 && b>0 && s>0, Limit[Simplify[s series], s -> 0] ]; lim[a, b] //TeXForm 

$\frac{\sqrt{\pi } e^{-\frac{b^2}{a}}}{\sqrt{a}}$

Summarizing, we have the limit:

$$\underset{s\to 0}{\text{lim}}\,s \vartheta _3(-b s,1-a s^2)=\frac{e^{-\frac{b^2}{a}} \sqrt{\pi }}{\sqrt{a}}$$

Now, the only nonzero term in the limit $s \to 0$ is then:

((a1 Sqrt[(d0 t)/l^2] EllipticTheta[3,-((d0 g π t)/(2 l)),E^(-((d0 π^2 t)/l^2))])/Sqrt[d0 t]) 

which takes the form:

c s EllipticTheta[3, -b s, 1 - a s^2] 

where:

c = a1; a = d0 π^2 t b = d0 g π t/2 

Hence, the symbolic limit is:

res = a1 lim[d0 π^2 t, d0 g π t/2]; res //TeXForm 

$\frac{\operatorname{a1} e^{-\frac{1}{4} \operatorname{d0} g^2 t}}{\sqrt{\pi } \sqrt{\operatorname{d0} t}}$

Let's check:

v = res /. {a1 -> 1, d0 -> 1, g -> 3/2, t -> 1/1000}; v //TeXForm N[v, 20] 

$\frac{10 \sqrt{\frac{10}{\pi }}}{e^{9/16000}}$

17.831208285385305584

in agreement with Bob's answer.