(\[CDidentity\]) we observe that $$\frac{1}{q_{m(n)}(x)}\left(S_{1/\rho_n, \mu_n,n}(x)-\frac{\left|\left|\widetilde{P}_{n}\right|\right|^2_{2, \rho_n d \mu/q_{m(n)}}}{\left|\left|q_{\mu_n, n-1}\right|\right|^2_{2, \mu_n}}S_{1/\rho_n, \rho_n d \mu/q_{m(n),n} }(x)\right)=\frac{q_{m(n)}^{-1}(x)}{\left|\left|q_{\mu_n,n-1}\right|\right|_{2,\mu_n}^2}$$ $$\times \int \left( \frac{q_{\mu_n,n}(x)q_{\mu_n,n-1}(t)-q_{\mu_n,n}(t)q_{\mu_n,n-1}(x)}{x-t}-\frac{P_{n}(x)q_{\mu_n,n-1}(t)-P_{n}(t)q_{\mu_n,n-1}(x)}{x-t}\right)$$ $$\times \frac{\mbox{d} \mu(t)}{q_{m(n)}(t)}$$ Let us consider the kernel $$\frac{q_{\mu_n,n}(x)q_{\mu_n,n-1}(t)-q_{\mu_n,n}(t)q_{\mu_n,n-1}(x)}{\left|\left|q_{\mu_n,n-1}\right|\right|_{2,\mu_n}^2q_{m(n)}(x)q_{m(n)}(t)(x-t)}-\frac{P_{n}(x)q_{\mu_n,n-1}(t)-P_{n}(t)q_{\mu_n,n-1}(x)}{\left|\left|q_{\mu_n,n-1}\right|\right|_{2,\mu_n}^2q_{m(n)}(x)q_{m(n)}(t)(x-t)}$$ $$=\frac{(q_{\mu_n,n}-P_n)(x)q_{\mu_n,n-1}(t)-(q_{\mu_n,n}-P_n)(t)q_{\mu_n,n-1}(x)}{\left|\left|q_{\mu_n,n-1}\right|\right|_{2,\mu_n}^2q_{m(n)}(x)q_{m(n)}(t)(x-t)}=\mathcal{K}(x,t).$$ From Taylor’s Theorem we obtain that $$\mathcal{K}(x,t)= \frac{1}{\left|\left|q_{\mu_n, n-1}\right|\right|^2_{2, \mu_n}}\left[\frac{(q_{\mu_n,n}-P_n)q_{\mu_n,n-1}}{q_{m(n)}^2}\right]^{\prime}(s)$$ for some $s$ in between of $x$ and $t$, so the assumption (\[extructura\]) completes the proof.

Asymptotic analysis {#asymptotic analysis}
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Let us consider the varying measure $\mu_n$ with $\displaystyle \mbox{d} \mu_n(x)/\mbox{d} x= (q_{m(n)}(x)\sqrt{1-x^2})^{-1}$, where $\displaystyle q_{m(n)}(x)= q_{\kappa}^{\displaystyle 2 \frac{1-a}{k} n}(x)=\left(\prod_{k=1}^{\kappa}(x-\zeta_j)\right)^{\displaystyle 2 \frac{1-a}{k} n}, \quad n\in \Lambda.$ Let $\sigma$ be the zero counting measure of $q_k$. This is $\displaystyle \sigma=\frac{1}{\kappa}\sum_{k=1}^{\kappa} \delta_{\zeta_k}.$ Set the analytic logarithmic potential corresponding to the measure $\sigma$: $$\label{externalfield}
g(z,{\sigma})= -\int \log\, (z-\zeta)\, \mbox{d}\, \sigma(\zeta).$$ We take the logarithmic branch such that $g(z,{\sigma})$ is analytic on a domain $D \subset K$ that contains the interval $[-1,\,1]$, and also for every $x \in [-1,\,1]$, $$\label{potentialsigma}
V^{\sigma}(x)= \int \log\, \frac{1}{|x-\zeta|} \mbox{d}\, \sigma(\zeta)=g(x,{\sigma})= -\int \log\, (z-\zeta)\, \mbox{d}\, \sigma(\zeta).$$ Since $\sigma$ is symmetric we $\displaystyle \int \arg (x-\zeta) \mbox{d}\, \sigma(\zeta)=0$. In each compact $K\subset D$ we have that $\displaystyle \frac{1}{2n} \log  \frac{1}{q_{m(n)}(z)}=(1-a) \,g(z,{\sigma})$ on $K.$

\[explicitexp\] Let $\displaystyle \mbox{d} \mu_n(x)/\mbox{d} x= (q_{m(n)}(x)\sqrt{1-x^2})^{-1}$, $n\in \mathbb{N}$ be a sequence of measures as above. Then 