case we see that setting $\beta=0$ in leads to an equation with no real solution for $(a,b)$. In this case the part $\beta\tfrac{a+b}{2ab}$ has non-zero limit when $\beta\to 0$. To be precise substitute $a=\beta a^\prime$ and $b=\beta b^\prime$ in , , and then we get $$\begin{aligned}
    1-\lambda+\alpha\beta\sqrt{a^\prime b^\prime}-\frac{a^\prime+b^\prime}{2a^\prime b^\prime}=&0\\
    1+\lambda+\frac{1}{\sqrt{a^\prime b^\prime}}-\alpha\beta\frac{a^\prime+b^\prime}{2}=&0.
    \end{aligned}$$ The above system is equivalent to the system , with $\alpha:=\alpha\beta$ and $\beta:=1$. If we set $\beta=0$ as in Case 1 we get $$\begin{aligned}
    1-\lambda-\frac{a^\prime+b^\prime}{2a^\prime b^\prime}=&0\\
    \label{Case2} 1+\lambda+\frac{1}{\sqrt{a^\prime b^\prime}}=&0.
    \end{aligned}$$ The above system has solution $0<a^\prime<b^\prime$ for $\lambda<-1$. Calculating the Jacobian we see that it is non-zero at $\beta=0$, so Implicit Function Theorem implies that $a^\prime$ and $b^\prime$ are continuous functions at $\beta=0$ in the case $\lambda<-1$.\
This implies that in the case $\lambda<-1$ the solutions of , are $a(\beta)= \beta a^\prime+o(\beta)$ and $b(\beta)= \beta b^\prime +o(\beta)$. Thus we have $$\begin{aligned}
    \lim_{\beta\to 0}\beta \gamma&=\frac{\alpha^2 ab+\tfrac{\beta^2}{ab}-2\alpha\beta (\tfrac{a+b}{\sqrt{ab}}-1)-(\lambda-1)^2}{4}\\&=\frac{\tfrac{1}{a^\prime b^\prime}-(\lambda-1)^2 }{4}=\frac{(\lambda+1)^2-(\lambda-1)^2 }{4}=\lambda,
    \end{aligned}$$ where in the equation one before the last we used .

**Case 3.** $|\lambda|< 1$\
Observe that neither nor has a real solution in the case $|\lambda|< 1$. This is because in this case asymptotically $a(\beta)=a^\prime \beta +o(\beta)$ and $b$ has a finite positive limit as $\beta\to0$. Similarly 