as in Case 2 let us substitute $a=\beta a^\prime $ in , , which gives $$\begin{aligned}
    1-\lambda+\alpha\sqrt{\beta a^\prime b}-\frac{\beta a^\prime+b}{2a^\prime b}&=0\\
    1+\lambda+\frac{\sqrt{\beta}}{a^\prime b}-\alpha\frac{\beta a^\prime+b}{2}&=0.
    \end{aligned}$$ If we set $\beta=0$ we get $$\begin{aligned}
    1-\lambda-\frac{1}{2a^\prime}&=0,\\
    1+\lambda-\alpha\frac{b}{2}&=0,
    \end{aligned}$$ which obviously has positive solution $(a^\prime,b)$ when $|\lambda|<1$. As before the Jacobian is non-zero at $\beta=0$, so $a^\prime$ and $b$ are continuous at $\beta=0$.\
Now we go back to the limit $\lim_{\beta\to 0}\beta\gamma$. We have $a(\beta)=\beta a^\prime+o(\beta)$, thus $$\begin{aligned}
    \lim_{\beta\to 0}\beta \gamma=\lim_{\beta\to 0}\frac{\alpha^2 ab+\tfrac{\beta^2}{ab}-2\alpha\beta (\tfrac{a+b}{\sqrt{ab}}-1)-(\lambda-1)^2}{4}=-\frac{(\lambda-1)^2}{4}.
    \end{aligned}$$

**Case 4.** $|\lambda|= 1$\
An analysis similar to the above cases shows that in the case $\lambda=1$ we have $a(\beta)=a^\prime \beta^{2/3}+o(\beta^{2/3})$ and $b$ has positive limit when $\beta\to 0$. In the case $\lambda=-1$ one gets $a(\beta)=a^\prime \beta+o(\beta)$ and $b(\beta)=b^\prime\beta^{1/3}+o(\beta^{1/3})$ as $\beta\to 0$.

Thus we can calculate the limit of $f_{\alpha,\beta,\lambda}$ as $\beta\to 0$ , $$\lim_{\beta\downarrow0} f_{\alpha,\beta,\lambda}(z) = 
\begin{cases}
(\alpha+(\lambda-1)z)^2, & \lambda>1, \\
 \alpha^2+(\lambda^2-1)\alpha z, & |\lambda| \leq 1, \\
 (\alpha-(\lambda+1)z)^2,  & \lambda<-1. 
 \end{cases}$$ The above allows us to calculate limiting $R$-transform and hence the Cauchy transform which implies .

Considering the continuous dependence of roots on parameters shows the following asymptotic behaviour of the double root $\delta<0$ and the simple root $\eta \geq \alpha$.

1.  If $|\lambda|>1$ then $\delta\to\alpha/(1-|\lambda|)$ 