Derivative of a limit of a sequence of polynomials equal to the limit of derivatives? Suppose that $f_1,f_2,\dots$ are polynomials of degree $m$ of a single variable and suppose $\lim_{n \to \infty} f_n(x) = f(x)$. Claim: $f'(x) = \lim_{n \to \infty} f_n'(x)$.
I know that it is not true for an arbitrary function. Here is a nice counterexample. However, I think that it works for polynomials. Is there a well-known result establishing this? My own thinking is this: the roots of polynomials are continuous in parameters and polynomials are fully determined by a constant and roots. This should imply that the result holds, but it seems like a clumsy and indirect way to go.
 A: You can show this simply by using polynomial interpolation. Choose $m+1$ points $x_0 < x_1 < \dots < x_m$ and let $P_i$ be the polynomial of degree $m$ for which $P_i(x_j) = \delta_{ij}, \, i,j = 0, \dots, m$. Then
$$
f_n(x) = \sum_j f_n(x_j)P_j(x), \quad f_n'(x) = \sum_j f_n(x_j)P_j'(x)
$$
and it follows that if $f_n(x_j) \to f(x_j)$ for all $j$, then 
$$
f_n'(x) \to f'(x)
$$
locally uniformly in $x$.
A: This answer is similar to Hans Engler's answer, in that it uses the idea of interpolating the polynomials at $m+1$ points to deduce something about the polynomials themselves.
You can say something stronger: the limit function $f$ is a polynomial of degree at most $m$ whose coefficients are the limits of the coefficients of $f_n$. To show this, write $f_n(x) = \sum\limits_{j=0}^{m}{a_{n}^{(j)}x^j}$, and choose $m+1$ points $x_0<x_1<\dots<x_m$. Let $y_n^{(i)} = f_n(x_i)$. For each $0\le i\le m$, we have $\sum\limits_{j=0}^{m}{a_{n}^{(j)}x_i^j} = f_n(x_i) = y_n^{(i)}$. We can express this as a matrix equation:
$$ \left(\begin{matrix} 1 & x_0 & x_0^2 & \dots & x_0^m \\ 1 & x_1 & x_1^2 & \dots & x_1^m \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_m & x_m^2 & \dots & x_m^m \end{matrix}\right)\left(\begin{matrix} a_n^{(0)} \\ a_n^{(1)} \\ \vdots \\ a_n^{(m)}\end{matrix}\right) = \left(\begin{matrix} y_n^{(0)} \\ y_n^{(1)} \\ \vdots \\ y_n^{(m)}\end{matrix}\right). $$
The square matrix on the LHS is the well-known Vandermonde matrix, which I will denote as $V$. Its determinant is $\prod\limits_{0\le i<j\le m}{(x_j-x_i)}$, and in particular it is invertible since all $x_i$ are distinct. So we can write
$$\left(\begin{matrix} a_n^{(0)} \\ a_n^{(1)} \\ \vdots \\ a_n^{(m)}\end{matrix}\right) = V^{-1}\left(\begin{matrix} y_n^{(0)} \\ y_n^{(1)} \\ \vdots \\ y_n^{(m)}\end{matrix}\right).$$
By hypothesis, $y_n^{(i)} = f_n(x_i)$ converges as $n\rightarrow\infty$ for each $i$, so the vector $\left(\begin{matrix} y_n^{(0)} \\ y_n^{(1)} \\ \vdots \\ y_n^{(m)}\end{matrix}\right)$ converges to some vector $\left(\begin{matrix} y^{(0)} \\ y^{(1)} \\ \vdots \\ y^{(m)}\end{matrix}\right)$. Thus $\left(\begin{matrix} a_n^{(0)} \\ a_n^{(1)} \\ \vdots \\ a_n^{(m)}\end{matrix}\right)$ converges to $\left(\begin{matrix} a^{(0)} \\ a^{(1)} \\ \vdots \\ a^{(m)}\end{matrix}\right) := V^{-1}\left(\begin{matrix} y^{(0)} \\ y^{(1)} \\ \vdots \\ y^{(m)}\end{matrix}\right)$, i.e. each coefficient $a_n^{(i)}$ converges to $a^{(i)}$. It follows that for every $x$ we have $$\lim\limits_{n\rightarrow\infty}{f_n(x)} = \lim\limits_{n\rightarrow\infty}{\sum\limits_{j=0}^{m}{a_n^{(j)}x^j}} = \sum\limits_{j=0}^{m}{x^j\left(\lim\limits_{n\rightarrow\infty}{a_n^{(j)}}\right)} = \sum\limits_{j=0}^{m}{a^{(j)}x^j}$$
and hence $f(x) = \sum\limits_{j=0}^{m}{a^{(j)}x^j}$. From here, the fact that $f_n'$ converges pointwise to $f'$ follows easily.
