When is it insufficient to treat the Dirac delta as an evaluation map? The Dirac delta "function" is often introduced as a limit of normal distributions $$\delta_a(x)=\frac{1}{a\sqrt{\pi}}e^{-\frac{x^2}{a^2}}\text{ as }a\to0^+.$$ Obviously, this sequence of functions converges to $0$ when $x\neq0$ and diverges otherwise. As far as I know, what is literally meant is that $$\int_{\mathbb{R}}f(x)\delta(x)dx\text{ is defined as }\lim_{a\to0^+}\int_{\mathbb{R}}f(x)\frac{1}{a\sqrt{\pi}}e^{-\frac{x^2}{a^2}}dx\text{, when $f$ is well behaved.}$$ I've read the wikipedia page on distributions and it was helpful. My question is this:
The Dirac delta acts on well behaved functions as the evaluation at zero map (and can be tweaked to evaluate at any point). I understand the intuitive motivation for using the (abuse of) notation "$\int_{\mathbb{R}}f(x)\delta(x) dx$" to represent a functional. Are there specific instances in mathematical analysis (broadly understood) where one needs to use the definition as "$\lim_{a\to0^+}\int_{\mathbb{R}}f(x)\frac{1}{a\sqrt{\pi}}e^{-\frac{x^2}{a^2}}dx$", and thinking of the Dirac delta as an evaluation map is insufficient?
 A: The delta function, as a generalised function, is the Fourier transform of the generalised function $1$,
$$ \int_\mathbb{R} 1 e^{-2\pi i kx} dx = \delta(k) \qquad \int_\mathbb{R} \delta(k) e^{2\pi i xk} dk =1 ,$$
this intuitively captures the fact that a wave localized in space is not localized in frequency and vice versa. I would have a hard time gaining that insight by thinking of the delta function as an evaluation map on a function space.

I agree that naively the notation suggests that these identities are impossible. You will notice that I was careful to say "the generalised function 1" not "the real number 1"; the distinction makes all the difference in the world. 
The generalised funciton $1$ is defined by the sequence of functions $\exp(-x^2/n^2)$. By definition the Fourier transform of a generalized function, $f(x)$, is the generalized function $g(k)$ defined by the sequence,
$$ g_n(k) = \int_\mathbb{R} e^{-2\pi i k x } f_n(x) dx $$
For the generalised function $1$ we have,
$$ g_n(k) = \int_\mathbb{R} e^{-2\pi i k x } e^{-x^2/n^2} dx $$
Evaluating this integral would produce a sequence of functions that are equivalent to the sequence defining the delta function. 
