The delta function (well, delta is not really a function; it is a distribution) can be defined as a limit of (among many other approximations) the following approximates of the unity:
the heat kernel:
$$\delta(x)=\lim_{\epsilon\rightarrow 0^{+}} \frac{e^{-\frac{x^2}{2\epsilon}}}{\sqrt{2\pi\epsilon}}\qquad (1)$$
or the Poisson kernel $$\delta(x)=\lim_{\epsilon\rightarrow 0^{+}} \frac{1}{\pi} \frac{\epsilon}{\epsilon^2+x^2}\qquad (2)$$
Check out the link
https://en.wikipedia.org/wiki/Dirac_delta_function#Representations_of_the_delta_function
Now, let $f$ be a tempered distribution. Can we interchange the limit with the integral so that
$$\int f(x)\delta(x)dx =\lim_{\epsilon\rightarrow 0^{+}}\int f(x) \frac{e^{-\frac{x^2}{2\epsilon}}}{\sqrt{2\pi\epsilon}}dx \qquad (*)$$
If $(*)$ is true, what is the justification ?
Notice that $(*)$ is true when $f$ is a smooth function with compact support, in which case both sides of $(*)$ give $f(0)$. With tempered distributions, we can not make sense of the pointwise value $f(0)$.
An easier question is when $f$ is a continuous oscillatory function, say $f(x)=e^{\dot{\imath}\alpha x^2}$, $\alpha \in \mathbb{C}$. What can we say then?
I think you're thinking about this the wrong way round. There isn't a function $\delta$ that is obtainable from kernels by pointwise convergence, and satisfies $\int_{\Bbb R}f(x)\delta (x) dx=f(0)$. Indeed, if we define $\delta$ as a pointwise limit, $\delta(x)=0$ for all $x\ne 0$ (while $\delta(0)=\infty$), so $\int_{\Bbb R}f(x)\delta (x) dx=0$. But what we can say is this: given a non-negative function $\phi(x)$ with $\phi(0)>0,\,\int_{\Bbb R}\phi(x) dx=1$, $$\lim_{\epsilon\to 0^+}\int_{\Bbb R}f(x)\tfrac{1}{\epsilon}\phi(\tfrac{x}{\epsilon})dx=\lim_{\epsilon\to 0^+}\int_{\Bbb R}f(\epsilon y)\phi(y)dy.$$If we can move the limit inside the integral, we obtain$$\int_{\Bbb R}\lim_{\epsilon\to 0^+}f(\epsilon y)\phi(y)dy=\int_{\Bbb R}f(0)\phi(y)dy=f(0).$$Note that this argument never tries to move the limit so as to write $$\int_{\Bbb R}f(x)\lim_{\epsilon\to 0^+}\tfrac{1}{\epsilon}\phi(\tfrac{x}{\epsilon})dx,$$which would be $0$ as aforesaid. The definition of the measure we call the Dirac delta is$$\int_a^b f(x)\delta(x)dx=\lim_{\epsilon\to 0^+}\int_a^bf(x)\tfrac{1}{\epsilon}\phi(\tfrac{x}{\epsilon})dx,$$not$$\delta(x)=\lim_{\epsilon\to 0^+}\tfrac{1}{\epsilon}\phi(\tfrac{x}{\epsilon}).$$The Dirac delta is not a function.