Dirac delta function $\delta(x)$ is defined with two properties:
1) At $x=0$ its value is $\infty$ and everywhere else it is $0$
2) Area under the curve is $1$
How does above definition result in $$\int\limits_{-\infty}^{\infty}\delta(x)f(x)dx = f(0) $$ ? Shouldn't the integral evaluate to $\delta(0)f(0)$?
I suggest reading about distributions (generalised functions), e.g. https://en.wikipedia.org/wiki/Distribution_(mathematics) . The point is, Dirac's $\delta$-function is not a function at all, at least not in the usual sense. What you have given as a "definition" is merely an intuition for it.
What $\delta$-function is is this: it is a linear functional that maps a certain class of functions (infintely smooth real functions that are nonzero only on a bounded set of $\mathbb R$) into real numbers, the following way:
$$\delta(f)=f(0)$$
Now, the right side looks like your right-hand side, but what about the left-hand side? One should note that, if $g:\mathbb R\to\mathbb{R}$ is a locally integrable function, then you can use it to define another functional of similar type:
$$T_g(f)=\int_{-\infty}^{\infty}g(x)f(x)dx$$
... and because for a given $T_g$ the function $g$ can be proven to be uniquely determined (up to a set of zero measure), then you can, in a sense, identify $T_g$ with $g$.
Now, in a common abuse of notation, we imagine that the functional $\delta$ is also identified with a fictional function $\delta(x)$ such that:
$$\delta(f)=\int_{-\infty}^{\infty}\delta(x)f(x)dx$$
so that is where the left-hand side is coming from. This is an abuse of notation because no such function $\delta(x)$ exists.