Iterated Limits Schizophrenia

That depends on what you mean by $\lim_{n\to \infty} g_n(x)$. The pointwise limit is zero, so the integral is zero. And if you put the parentheses so: $\lim_{n\to \infty} (g_n(x))$ then we have to conclude that the limit is zero and $$\int_0^1 \lim_{n\to \infty} (g_n(x))\,dx=\int_0^1 0\,dx=0$$

However, you can also read this as $\lim_{n\to \infty} (g_n\, dx)$ where $dx$ denotes the usual Lebesgue measure. This is a limit of measures and it does, in fact, converge to the point measure at $0$, so that $$\int_{[0,1]} \lim_{n\to \infty} (g_n\, dx)=\int_{[0,1]}d\delta_0=1$$ (I've changed the notation because $\int_0^1$ can be confusing in this case, as $\int_{(0,1)}d\delta_0=0$).

So the "paradox" actually comes from different kinds of limits, and the arising confusion comes from not explaining the kind of convergence that is used in each case.

The equality $\lim_{n\to\infty}g_n(x)=\delta(x)$ is incorrect. In fact, $\lim_{n\to\infty}g_n(x)=0$ for all $x$ (including $x=0$). For $D_n$, $$ \lim_{n\to\infty}D_n(x)=\begin{cases}0,&\ x\ne0,\\ \infty,&\ x=0\end{cases} $$ So, saying that the two limits are equal makes no sense.

The physicists can "get away with it" because they formally treat $\delta$ as a function, although it is not. This is well-understood by mathematicians. One uses the functions $g_n$, $D_n$ to define linear functionals over the space of continuous functions. Given a continuous function $f$, one defines $$ \varphi_n:f\mapsto\int_0^1f(t)g_n(t)dt. $$ One can check that $\lim_{n\to\infty}\varphi_n(f)=f(0)$. So the functionals $\varphi_n$ converge pointwise to the functional $\delta:f\mapsto f(0)$. Physicists love an intuitive vision of things, so they claim that this new functional should also be integration against a function, that they call $\delta(x)$, the Dirac function. So, for them, $$\tag{1} \delta(f)=\int_0^1f(t)\delta(t)dt. $$ This last equality makes no sense mathematically. What happens is that one wants to think of $\delta(t)dt$ as a measure, and integrate $f$ against it. And this can be done: $$\tag{2} \delta(f)=\int_{[0,1]}f(t)d\delta(t), $$ where now one considers the measure $\delta$ given by $$ \delta(S)=\begin{cases}1,&\ 0\in S,\\ 0,&\ 0\not\in S\end{cases} $$ What happens then is that physicists use $(1)$, while they should be using $(2)$. But they use $(1)$ in a way that it just becomes notation for $(2)$ which is the right one mathematically.