Can the Dirac Delta, $\delta $, be obtained by taking the limits of a rectangular pulse? Can the Dirac Delta, $\delta $, be obtained by taking the limits of a rectangular pulse of width w and amplitude 1/w (ie. as w tends to zero)? How does the results differ from using the Gaussian, the sinc function or other commonly used function?.
 A: The as-usual-unhelpful response is "yes". :)
But, yes, Dirac's distribution is indeed the limit that you describe (oop, a person has to worry "in what topology"?... but let's delay that...)
Probably the real question is something subsequent/pursuant to the literal question you've asked...
... so perhaps either refinement/editing of the question, or a different question, would better serve "the questioner's" interests...
Clarify?
In response to follow-up: if the only thing you'll do with those rectangular pulses is integrate them against continuous functions $f$, you will reliably get $f(0)$ in the limit.
However, in contrast to using Gaussians or other smoother things, it is maybe less intuitive how to take a derivative of those rectangular pulses. Not impossible, by any means, but less intuitive. In fact, if one already knows how to take derivatives of distributions, then one correctly finds that the derivative of the $w$-th rectangular pulse is something like $2(\delta_{-w}-\delta_w)/w$, which does indeed give $\delta'_0$ in the limit $w\to 0$... but maybe this is harder to see than the analogue for Gaussians, where the derivatives of the things along the way are classical rather than distributional derivatives.
A: I'm putting this as an answer as I can't really fit this in a comment.
Let $R_n(x)$ denote a function which is $n$ on the interval $[-1/2n,1/2n]$ and zero elsewhere. We can prove that $\lim_{n\rightarrow\infty}\int R_n(x)f(x)=f(0)$. To see this, note that:
$$f(0)=\int_{-1/2n}^{1/2n}nf(0)$$
Thus
$$\left|\int R_n(x)f(x) - f(0)\right|=n\left|\int_{-1/2n}^{1/2n}f(x)-f(0)\right|\le n\int_{-1/2n}^{+1/2n}|f(x)-f(0)|$$
Assuming $f$ is continuous (test functions are in fact uniformly continuous), we can choose an $n$ such that the integrand is less than arbitrary $\epsilon$, giving the desired result.
To be honest though distributions are complicated, so I am not sure if this is sufficient. I would think that using a limit of Guassians would simplify some problems, as perhaps you can differentiate and use integration by parts in the immediate steps.
