of the algebraic equation $$\label{eq:curveN}
\begin{split}
za\xi^3&+(A_2z+B_2^N)\xi^2+(z+B_1^N)\xi+1=0, \\
B_2^N&=a(1-c_N),\quad B_1^N=1-c_N(1-\beta_N)+a(1-c_N\beta_N).
\end{split}$$ that behaves as $-\frac{1}{z}$ as $z\rightarrow\infty$. If we denote by $\xi_j^N(z)$ the solutions of (\[eq:curveN\]) with asymptotic behavior (\[eq:xiinfty\]), but with $c$ and $\beta$ replaced by $c_N$ and $\beta_N$, then we have $\xi_1^N(z)=m_N(z)$. In [@CS], it was shown that the measure $\hat{F}_N$ has a continuous density on $\mathbb{R}_+$ and a point mass of size $1-c_N$ at $0$ and hence $\xi_1^N(z)$ also have the asymptotic behavior (\[eq:xizero\]) near $z=0$ with $c$ replaced by $c_N$. By Remark \[re:baik\], all the results in the previous section will remain valid for (\[eq:curveN\]) and $\xi_j^N(z)$, with $\beta$ and $c$ replaced by $\beta_N$ and $c_N$. We will denote the Riemann surface defined by (\[eq:curveN\]) ${\mathcal{L}}_N$.

Let $\Delta_N$ be the determinant of the quartic polynomial (\[eq:quart\]) $$\label{eq:quartN}
\begin{split}
a^2(1-c_N)\xi^4
&+2(a^2(1-c_N\beta_N)+a(1-c_N(1-\beta_N))\xi^3\\
&+(1-c_N(1-\beta_N)+a^2(1-c_N\beta_N)+4a)\xi^2 +2(1+a)\xi+1=0
\end{split}$$ Since the determinant $\Delta$ of the quartic polynomial (\[eq:quart\]) is continuous in the parameters $a$, $\beta$ and $c$, for large enough $N$, $M$ and $N_1$, we can assume that the determinant $\Delta_N<0$. Then let $\gamma_1^N<\gamma_2^N$ be the real roots of (\[eq:quartN\]) and $\gamma_{3}^N$ and $\gamma_4^N$ be the complex conjugate roots of (\[eq:quartN\]) and let $\lambda_1^N<\lambda_2^N$ and $\lambda_{3}^N$, $\lambda_4^N$ be their images under the 