Define the sequence of functions $$\delta_n(x) = \begin{cases} n \quad & \frac{-1}{2n} < x < \frac{1}{2n} \\0 & \text{else} \end{cases}$$ and define the polynomial $g(x)=c_1 + c_2x + \dots c_kx^k$. Then $$\lim_{n \to \infty} \int_{-1}^{1}g(x)\delta_n(x)=\lim_{n \to \infty} (c_1+c_3\frac{2}{3\cdot4}+\dots c_k\frac{2}{k\cdot n^k}=c_1=g(0).$$
So this is just exactly what would happen if I was allowed to take the limit inside the integral so that $$\int_{-1}^{1}g(x)\delta(x)=g(0),$$ but I am not allowed to take the limit inside because the sequence of functions doesn't converge uniformly, right? So is this just a coincidence, or am I missing something?
The rigorous way to justify "taking limit inside the integral" to a non-function $\delta (x)$ is as follows:
$$ \int_{-1}^1 g(x) \delta_n (x) = n\int_{-1/2n}^{1/2n} g(x) dx = g(0) + \int_{-1/2n}^{1/2n} (g(x)-g(0)), $$
while
$$|g(x)-g(0)| = |c_2 x + \cdots c_k x^k| \le (|c_2| + \cdots |c_k|)|x| \le (|c_2| + \cdots |c_k|).$$
Thus
$$\left| \int_{-1/2n}^{1/2n} (g(x)-g(0))dx\right| \le \frac{(|c_2| + \cdots |c_k|)}{n} \to 0$$
as $n\to \infty$.
In general you have
$$ \int_{-1}^1 g(x) \delta_n(x) dx\to g(0)$$
for all $g$ continuous at $0$, since one can bound $g(x)-g(0)$.