Delta function representation Suppose I have a set of functions $(f_\epsilon)$ such that as $\epsilon\to 0$, $f_\epsilon\to F$ s.t. 
$F(x)=0$ for $x\neq 0$ and $F(x)=\infty$ for $x=0$;
$\int_{-\infty}^\infty f_\epsilon(x) dx=1$ for all $\epsilon >0$
Then can I conclude that the limit is the delta function $\delta(x)$? (which has the sampling property too)
Thanks
 A: To be able to answer this question, you have to define a few important concepts such as distributions or generalized functions, the limiting process etc. and you have to be extremely careful when performing the limiting step. The usual intuitions generally (and only formally) fail to make sense. The main problem is that the Dirac Delta does not mean anything outside an integral,sometimes even inside, in particular it is not defined pointwise!! Therefore, sampling with the Dirac comb, which is a function(!) consisting Dirac Deltas repeated periodically is not a valid operation if we write
$$
u(k) = \operatorname{comb}(t)u(t)
$$
where $k\in\mathbb{Z}$ and $t\in\mathbb{R}$.
In other words, there is no pointwise multiplication that gives you the sampled version of $u(t)$ but it is, without a doubt, a neat shortcut to state the operation. 
For a wonderful storyline I strongly recommend you the Lecture 12 of Brad Osgood's Fourier Transform and Applications course. He defines almost all practical definitions and mention how they fail (roughly though). But the course material spends quite some effort on this issue. 
