While I was reading more about the Dirac delta, I came across the following answer: Answer about Dirac Delta
As such, my question is what are the differences (either in their use or in their results) between specifying $f(0)$ and $ \delta (x) f(x)$? I know $ \delta (x)$ is a functional while $f(0)$ is a value, but what does this entail in practice?
One reason I believe could be the preservation of the domain of $f(x)$, but I do not believe this is the only (or correct) reason.
As the link mentioned, $\delta(x)$ is not a function but a functional (generalized function or distribution). Let $C^\infty_c(\mathbb R)$ be the space of compactly supported smooth functions (this space of so-called "test" functions enjoy some nice properties but not essential for the current discussion), then technically $\delta$ is a linear functional from $C^\infty_c(\mathbb R)$ to $\mathbb R$ such that $\delta(f)=f(0)$.
$\delta$ is called a generalized function for good reason. Given a usual function $g$ (that is at least locally integrable), then we can define a functional on $C^\infty_c(\mathbb R)$ to $\mathbb R$, defined by $$f\mapsto \int_{-\infty}^{\infty} g(x)f(x)\,dx$$
Although this definition only makes sense for usual funcitons, we often use the same notation for a genearalized function such as the Dirac delta: $$f\mapsto \int_{-\infty}^{\infty} \delta(x)f(x)\,dx = f(0)$$
So it's just a matter of notation. To make sense of the multiplication $\delta(x)f(x)$ is a much more complicated issue. It's better to think we have never formed $\delta(x)f(x)$ but the operator $\int_{-\infty}^{\infty}\delta(x)\{\cdot\}\,dx$ acts on $f(x)$, just like $\int_{-\infty}^{\infty} g(x)\{\cdot\}\,dx$ acts on $f(x)$ (though in the latter case, $f(x)g(x)$ is indeed legit and necessary.).