  For $a\in (0,1)$ let $L^k_m(a)$ be the error of the best approximation of the
function $\sgn(x)$ on the two symmetric intervals $[-1,-a]\cup[a,1]$ by
rational functions with the only possible poles of degree $2k-1$ at the origin
and of $2m-1$ at infinity. Then the following limit exists \begin{equation}
\lim_{m\to \infty}L^k_m(a)(\frac{1+a}{1-a})^{m-{1/2}} (2m-1)^{k+{1/2}}=\frac 2
\pi(\frac{1-a^2}{2a})^{k+{1/2}} \Gamma(k+\frac 1 2). \end{equation}
