Let $f \in S(\mathbb{R}^n)$ is it true that $$\frac{1}{(2\pi)^n} \lim_{\epsilon \rightarrow 0} \sum_{z\in \mathbb{}{Z^n}} \int_\mathbb{R^n} f\left( \frac{x}{\epsilon} \right) e^{iz (x-a\epsilon)} dx = f(a)$$
Formally, if we put the sum inside and use the formal series representation of dirac delta $\delta (x-a) = \frac{1}{(2\pi)^n}\sum_{z\in \mathbb{}{Z^n}} e^{iz (x-a)}$ then it seems we get the result. But this is not rigorous.
The integral equals $\epsilon^n\widehat{f}(\epsilon z) e^{-ia\epsilon z}$, and thus the sum can be viewed as a Riemann sum for $\int_{\mathbb R^n}\widehat{f}(t)e^{-iat}\, dt=f(a)$.