Lagrange interpolation as two points get closer and closer Suppose, you want to perform a Lagrange interpolation for $x_0$, $x_0+\varepsilon$ and $x_1$. What happens as $\varepsilon \to 0$?
Here are the Lagrange-polynomials
\begin{gather*}
L_0(x)=\frac{(x-(x_0+\varepsilon))(x-x_1)}{(x_0-(x_0+\varepsilon))(x_0-x_1)}=\frac{(x-(x_0+\varepsilon))(x-x_1)}{-\varepsilon (x_0-x_1)} \\
L_1(x)=\frac{(x-x_0)(x-x_1)}{\varepsilon(x_0+\varepsilon -x_1)} \\
L_2(x)=\frac{(x-(x_0+\varepsilon))(x-x_0)}{(x_1-(x_0+\varepsilon))(x_1-x_0)}
\end{gather*}
We then have $P(x)=y_0L_0(x)+y_1L_1(x)+y_2L_2(x)$. What happens as $\varepsilon \to 0$? What's the degree of the resulting polynomial?
 A: I assume that we wish to interpolate a function $f : \mathbb{R} \rightarrow \mathbb{R}$ at the nodes $x_0,x_0+h,x_1$ and that the goal is to determine how the interpolating polynomials behave when $h$ tends to zero.

The Newton form of the interpolation polynomial is 
$$p_h(x) = f(x_0) + f[x_0,x_0+h](x-x_0)+f[x_0,x_0+h,x_1](x-x_0)(x-x_0-h)$$
where the relevant divided differences are given by
$$ f[x_0,x_0+h,x_1] = \frac{f[x_0+h,x_1]-f[x_0,x_0+h]}{x_1-x_0}$$
and
$$ f[x_0+h,x_1] = \frac{f(x_1)-f(x_0+h)}{x_1-x_0-h}, \quad f[x_0,x_0+h] = \frac{f(x_0+h)-f(x_0)}{h}$$
Now let $x$ be fixed. Our concern is the behavior of $p_h(x)$ as $h$ tends to zero. 
If $f$ is continuous at $x_0$, then
$$ f[x_0+h,x_1] \rightarrow f[x_0,x_1], \quad h \rightarrow 0, \quad h \not = 0.$$
If $f$ is differentiable at $x_0$, then
$$ f[x_0,x_0+h] \rightarrow f'(x_0)  \quad h \rightarrow 0, \quad h \not = 0.$$
We conclude that if $f$ is differentiable at $x=x_0$, then 
$$p_h(x) \rightarrow p(x) \quad h \rightarrow 0, \quad h \not = 0$$
where
$$ p(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f[x_0,x_1] - f'(x_0)}{x_1-x_0}(x-x_0)^2.$$
We observe that this polynomial has degree at most $2$ and satisfies
$$ p(x_0) = f(x_0)$$
and
$$ \begin{align} p(x_1) &= f(x_0) + f'(x_0)(x_1-x_0) + (f[x_0,x_1]-f'(x_0))(x_1-x_0) \\ & = f(x_0) + \frac{f[x_1]-f[x_0]}{x_1-x_0}(x_1-x_0) = f(x_1) \end{align}$$
as expected. Moreover, the derivative of $p$ is given by
$$ p'(x) = f'(x_0) + 2 \frac{f[x_0,x_1]-f'(x_0)}{x_1-x_0}(x-x_0)$$
It follows that
$$ p'(x_0) = f'(x_0).$$
We conclude that $p$ interpolates $f$ at $x_0$ and $x_1$ and $p'$ interpolates $f'$ at $x_0$.
