[@HP00] for the Wishart matrix).

We will find the limit by calculating limits of the $R$-transform, since convergence of the $R$-transform implies weak convergence. Observe that from Remark \[rem:F\_lambda\_sym\] we can consider only $\lambda\geq0$, however we decided to present all cases, as the consideration will give asymptotic behaviour of support of fGIG measure. In view of , the only non-trivial part is limits of $\beta\gamma$ when $\beta\to0$. Observe that if we define $F(a,b,\alpha,\beta,\lambda)$ by $$\begin{aligned}
    \left(1-\lambda+\alpha\sqrt{ab}-\beta\frac{a+b}{2ab},1+\lambda+\frac{\beta}{\sqrt{ab}}-\alpha\frac{a+b}{2}\right)^T &=\left(f(a,b,\alpha,\beta,\lambda),g((a,b,\alpha,\beta,\lambda))\right)^T\\
    &=F(a,b,\alpha,\beta,\lambda). 
    \end{aligned}$$ Then the solution to the system , are functions $(a(\alpha,\beta,\lambda),b(\alpha,\beta,\lambda))$, such that $F(a(\alpha,\beta,\lambda),b(\alpha,\beta,\lambda),\alpha,\beta,\lambda)=(0,0)$. To use Implicit Function Theorem, by calculating the Jacobian with respect to $(a,b)$, we observe that $a(\alpha,\beta,\lambda)$ and $b(\alpha,\beta,\lambda)$ are continuous (even differentiable) functions of $\alpha,\beta>0$ and $\lambda\in \mathbb{R}$.

**Case 1.** $\lambda>1$\
Observe if we take $\beta=0$ then a real solution $0<a<b$ for the system , $$\begin{aligned}
\label{Case1}
    1-\lambda+\alpha\sqrt{ab}&=0\\
    1+\lambda-\alpha\frac{a+b}{2}&=0
    \end{aligned}$$ still exists. Moreover, because at $\beta=0$ Jacobian is non-zero, Implicit Function Theorem says that solutions are continuous at $\beta=0$. Thus using we get $$\begin{aligned}
    \beta \gamma=\frac{\alpha^2 ab+\tfrac{\beta^2}{ab}-2\alpha\beta (\tfrac{a+b}{\sqrt{ab}}-1)-(\lambda-1)^2}{4}=\frac{\tfrac{\beta^2}{ab}-2\alpha\beta (\tfrac{a+b}{\sqrt{ab}}-1)}{4}.
    \end{aligned}$$ The above implies that $\beta\gamma\to 0$ when $\beta\to 0$ since $a,b$ have finite and non-zero limit when $\beta\to 0$, as explained above.

**Case 2.** $\lambda<-1$\
In that 