In my research (in communication theory) I ran across the limit of the confluent Lauricella function of three variable, i.e. $$\lim_{x\to\infty}\Phi_2^{(3)}\left(1,a+x,-x;c+1;t,t\frac{1}{1+b_1},t\frac{1}{1+b_1\frac{x}{(x-b_2)}}\right)$$ I've tried to use the series representation: $$\Phi_2^{(3)}\left(1,a+x,-x;c+1;t,t\frac{1}{1+b_1},t\frac{1}{1+b_1\frac{x}{(x-b_2)}}\right)=\sum_{m_1,m_2,m_3}\frac{(1)_{m_1}(a+x)_{m_2}(-x)_{m_3}}{(c+1)_{m_1+m_2+m_3}m_1!m_2!m_3!}t^{m_1}\left(\frac{t}{1+b_1}\right)^{m_2}\left(\frac{t}{1+\frac{b_1x}{(x-b_2)}}\right)^{m_3}$$ If $t<1$ then I can use this representation. Taking the limit of the last multiplier, but still I have to deal with the limit of the Pochhammer functions. Moreover, if $t>1$ (and this is the case of primary importance) this fails at all.
It was tempting to use limiting relation with Lauricella $F_D^{(3)}$ function: $$\Phi_2^{(3)}\left(1,a+x,-x;c+1;t,t\frac{1}{1+b_1},t\frac{1}{1+b_1\frac{x}{(x-b_2)}}\right)=\lim_{\epsilon\to \infty}F_D^{(3)}\left(\epsilon,1,a+x,-x;c+1;\frac{t}{\epsilon},\frac{t}{\epsilon}\frac{1}{1+b_1},\frac{t}{\epsilon}\frac{1}{1+b_1\frac{x}{(x-b_2)}}\right)$$ and then use its integral representation (over the range $0$ to $1$) (in this case the terms with $x$ cancell out), but the problem is, that it is valid only if $c+1>\epsilon$ and this is false. I've tried to represent $\Phi_2^{(3)}$ in terms of Laplace transform, take the limit and getting some ugly exponent, expanding this exponent term in series and backwards collect the Lauricella function. But numeric simulation demonstrated that this approach does not work.
So my question is: is there any way to find the initial limit.