What's the point of asymptotics? I presume this is a rather stupid question, but I hope some of you can find a bit of time to entertain it.
Looking at asymptotic behavior of estimators/test statistics etc means looking at their behavior as sample size approaches infinity. But if our sample size is infinite, doesn't that mean that we already have the whole population?
Case in point: consistency of OLS estimates. Under certain assumptions we are guaranteed convergence of estimates to true parameters in the limit, but at infinite sample size we already have the whole population, so why would we even want to estimate anything when every possible data point is already contained in the sample, all we have to do is to perform a lookup.

Most asymptotic results are closely connected to finite first-order results. We'll review this in the non-probabilistic case and then extend to the probabilistic case.
Non random case: reduced order analysis
Recall the Taylor series of the function $\sin$ around the point $x=0$:
\begin{equation*}
 \sin x - 0 = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots,
\end{equation*}
which most learn about in introductory calculus courses. The series helps us understand the behavior of the function $\sin$ function near $x=0$. The plot below (from Wikipedia) shows that these four terms largely reproduce the behavior of the curve near $x=0$:

Thus, to study the behavior of $\sin(x)$ near $x=0$, very little is lost in simply studying the behavior of the polynomial $x \mapsto x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}$. Of course, the polynomial is much more susceptible to study, so this is convenient.
Non random case: first order analysis
The close relationship between the function $\sin$ and its truncated Taylor series also holds in the first order, when we approximate
\begin{equation*}
 \sin x - 0 = x - \dots,
\end{equation*}
For example, most calculus students have shown that
\begin{equation*}
 \lim_{x \to 0} \frac{\sin x}{x} = 1.
\end{equation*}
This result can be interpreted as saying that the function $\sin$ and the identity function $x \mapsto x$ are indistinguishable near $x=0$.
We stress that the limit result, although asymptotic, is merely reflecting our knowledge gained from the figure above: the two functions are very close near $x=0$.
Probabilistic case
There is an extension to Taylor series which is designed for estimators rather than functions. We'll summarise the extension and illustrate how it is behind most "asymptotic" results.
Let $P_0$ denote the true (unknown) distribution, let $\theta(P_0)$ denote the parameter of interest, and let $T(X_1, \dots, X_n)$ be the estimator using i.i.d. data $X_i \sim P_0$. Then, the following expansion is a first order expansion of the estimator around the parameter:
\begin{align*}
 T(X_1, \dots, X_n) - \theta(P) = \frac{1}{n} \sum_{i=1}^n \varphi(X_i; P_0) + \dots.
\end{align*}
The term on the left is the estimator error, i.e. the difference between the estimator and its target (the parameter). The term on the right is a first order expansion, analogous to $x$ in the $\sin$ case. The function $\varphi$ is called the "influence function" and determines the asymptotic behavior, along with the remainder.
(More details later..)
