Does this sequence of polynomials have a limit? Consider the sequence of polynomials $p_n$ defined as follows: $p_n$ is the unique polynomial of degree $2n+1$ satisfying
$$p_n(0) = 0$$
$$p_n(1) = 1$$
$$p_n^{(k)}(0) = p_n^{(k)}(1) = 0 \text{ for $k=1$ to $n$}$$
where $p_n^{(k)}$ is the $k$th derivative of $p_n$. For example,
$$p_0(x)=x$$
$$p_1(x)=-2 x^3+3 x^2$$
$$p_2(x)=6 x^5-15 x^4+10 x^3$$
$$p_3(x)=-20 x^7+70 x^6-84 x^5+35 x^4$$
$$p_4(x)=70 x^9-315 x^8+540 x^7-420 x^6+126 x^5$$
$$p_5(x)=-252 x^{11}+1386 x^{10}-3080 x^9+3465 x^8-1980 x^7+462 x^6$$
To give some intuition, here is an animated plot of $p_n$ for $n=0$ to $50$.

How can I determine, with proof, $\lim_{n \to \infty} p_n(x)$ (if it exists)? I've never worked with functions implicitly defined in this way before, and I have no idea where to begin.
 A: The functions which are $f(-1)=-1,f(1)=1$ and all other derivatives are zero are
$$C_n\int_0^x (1-t^2)^n dt$$  
It follows that, pointwise, they reach constants as $n\to\infty$ because the bulk of $(1-t^2)^n$ narrows towards $x=0$.
The OP's functions are 
$$D_n\int_0^x (t-t^2)^n \, dt$$  
Let $I_n=\int_0^1(t-t^2)^n \, dt=1/D_n$.  Apply integration by parts, but let the integral of $1$ be $t-1/2$ instead of $t$, to use the symmetry between $0$ and $1$.
$$
\begin{align}
& I_n=\left.(t-1/2)(t-t^2)^n\vphantom{\frac11}\right|_0^1-\int_0^1(t-1/2)n(t-t^2)^{n-1}(1-2t) \, dt\\
= {} & 0-n\int_0^1(-2t^2+2t-1/2)(t-t^2)^{n-1}dt\\
= {} & -2nI_n+n/2I_{n-1}\\[6pt]
& (1+2n)I_n=(n/2)I_{n-1}\\[6pt]
& (2n+1)D_{n-1}=(n/2)D_n\\[6pt]
& (2n+1)(2n)D_{n-1}=n^2D_n
\end{align}
$$
Now David's formula for $D_n$ follows by induction.
A: @Michael has already found a very elegant answer, but I would like to contribute an alternate approach that I stumbled across. It turns out that the polynomials $p_n$ are precisely the degree $2n+1$ Bernstein polynomial approximants of $u(x-1/2)$, where $u$ is the unit step function (and $u(0) = 1/2$). These are well-known to converge to the original function.
