We interpolate $f(x) = \frac{x^2}{\sin(\arccos(x))^2} = \frac{x^2}{1-x^2}$ on the Chebyshev-nodes in the $(-1,1)$ interval.

**Lemma 1.** [1, 6.4 lemma]. *If $x_1, x_2, \ldots, x_{n+1}$ are the roots of the polynomial $\omega_{n+1}(x)$, then the $\ell_i(x)$ Lagrange-polynomials can be written in the following form
$$\ell_i(x) = \frac{\omega_{n+1}(x)}{(x-x_i)\cdot \omega'_{n+1}(x_i)}.$$*
We also need a trigonometric identity.

**Proposition 1.** *Let $n$ be an arbitrary positive, even integer, then the following holds
$$\sum^{n-1}_{k=0} (-1)^k \cot\left(\frac{(2k+1) \pi}{2n}\right) = n.$$*
**Proof.** A similar statement can be found in [2, 141. b.], that is,
$$\sum^{n-1}_{k=0} (-1)^k \cot\left(\frac{(2k+1) \pi}{4n}\right) = n.$$
Using that $\cot(2x)=\frac{1}{2}(\cot(x) - \tan(x))$ holds for all $x$, it is sufficient to show that
$$\sum^{n-1}_{k=0} (-1)^k \tan\left(\frac{(2k+1) \pi}{4n}\right) = -n,$$
from which follows the statement.

I prove the last inequality based on the solution of [2, 141. b.].
We know [2, 132. c] that
$$\tan(n\alpha) = \frac{\binom{n}{1} \tan \alpha - \binom{n}{3} \tan^3 \alpha + \binom{n}{5} \tan^5 \alpha - \ldots}{1 - \binom{n}{2} \tan^2 \alpha + \binom{n}{4} \tan^4 \alpha - \ldots}.$$
If $\alpha$ is one of the following numbers $\frac{\pi}{4n}, \frac{5\pi}{4n}, \ldots, \frac{(4n-3)\pi}{4n}$, then $\tan(n\alpha) = 1$, thus the $\alpha$ numbers give all of the roots of the following polynomial
$$1 - \binom{n}{1} x - \binom{n}{2} x^2 + \binom{n}{3}x^3 + \binom{n}{4}x^4 - \binom{n}{5}x^5 - \ldots - \binom{n}{2k-1} x^{2k-1} - \binom{n}{2k} x^{2k}$$
where $n = 2k$ and $n \equiv 2 \pmod{4}$. If $n \equiv 0 \pmod{4}$, then
$$1 - \binom{n}{1} x - \binom{n}{2} x^2 + \binom{n}{3}x^3 + \binom{n}{4}x^4 - \binom{n}{5}x^5 - \ldots + \binom{n}{2k-1} x^{2k-1} + \binom{n}{2k} x^{2k}$$
we get the roots of the polynomial above.

Notice that $\tan\left( \frac{(4n-3)\pi}{4n} \right) = -\tan\left( \frac{3\pi}{4n} \right), \tan\left( \frac{(4n-7)\pi}{4n} \right) = -\tan\left( \frac{7\pi}{4n} \right), \ldots, \tan\left( \frac{(2n+1)\pi}{4n} \right) = -\tan\left( \frac{(2n-1)\pi}{4n} \right),$ that is, the sum of the roots is
$\sum^{n-1}_{k=0} (-1)^k \cot\left(\frac{(2k+1) \pi}{2n}\right),$
and applying Viète-formula we have
$$-\frac{\binom{n}{2k-1}}{\binom{n}{2k}} = -\frac{n}{1} = -n.$$
**Proposition 2.** *Let $L_{n-1}$ be the $n$th interpolating polynomial on the Chebyshev-nodes of $f$, then $L_{n-1}(0) = -\cos(\frac{\pi}{2} \cdot n)$ for all $n \in \mathbb{N}$.*

**Proof.** By definition
\begin{equation}L_{n-1}(x) = \sum^{n-1}_{k=0} \ell_k(x) f(x_k),
\end{equation}
where $x_k = \cos \left( \frac{(2k+1)\pi}{2n} \right), k = 0,1,\ldots,n-1$. By Lemma 1. we can easily calculate the Lagrange-polynomials at $0$. Since $T_n'(x) = \frac{n\cdot \sin(n \cdot \arccos(x))}{\sqrt{1-x^2}}$, we have that
$$T_n'(x_k) = \frac{n\cdot\sin\left(\frac{2k+1}{2}\pi \right)}{\sin\left(\frac{2k+1}{2n}\pi \right)}.$$
Furthermore $T_n(0) = \cos(\frac{\pi}{2}n)$, thus
$$\ell_k(0) = \frac{\sin\left(\frac{2k+1}{2n}\pi \right) \cos(\frac{\pi}{2}n)}{-n\cdot\sin\left(\frac{2k+1}{2}\pi \right) \cos \left( \frac{(2k+1)\pi}{2n} \right)}.$$
Now, we calculate $f(x_k)$:
$$f(x_k) = \frac{\cos \left( \frac{(2k+1)\pi}{2n} \right)^2}{\sin \left( \frac{(2k+1)\pi}{2n} \right)^2}.$$
Since $\sin\left(\frac{2k+1}{2}\pi \right) = (-1)^k$, we get that
$$ \ell_k(0)f(x_k) = (-1)^{k+1} \frac{\cot \left( \frac{(2k+1)\pi}{2n} \right) }{n}.$$
From which follows that
$$L_{n-1}(0) = \frac{-1 \cdot \cos(\frac{\pi}{2}n)}{n} \sum^{n-1}_{k=0} (-1)^{k} \cot \left( \frac{(2k+1)\pi}{2n} \right),$$
however if $n$ is even by Proposition 1. we have
$$
\sum^{n-1}_{k=0} (-1)^{k} \cot \left( \frac{(2k+1)\pi}{2n} \right) = n,
$$ and we obtain what we need.

If $n$ is odd, it is true that $L_{n-1}(0) = f(0) = 0$, since $0$ is a root of the $n$th Chebyshev-polynomial. Since $T_n(0) = \cos(\frac{\pi}{2} \cdot n) = 0$, it is sufficient to take that term of (1) when $x_k = 0$. Then
$$f(x_k) \frac{T_n(x)}{x_k T_n'(x_k)} = f(x_k) = 0,$$
since $\frac{T_n(x)}{xT_n'(x)} = 1$ if $x = 0$ and this is a root of $T_n(x)$.

Therefore, we get that $L_{n}(0) \not\rightarrow f(0)$ when $n \rightarrow \infty$.

**References.**

[1] J.C. Mason, David C. Handscomb, *Chebyshev Polynomials*, CRC Press, 2002.

[2] A. M. Yaglom, I. M. Yaglom *Challenging Mathematical Problems With Elementary Solutions, Vol. 2*, Courier Corporation, 1987