Let me show the supposition extracted from S.-P. Zhu et al. / Computers and Mathematics with Applications 64 (2012),
Consider the system of coupled Black-Scholes equations (Regime-switching) for the value of a European put option: $$ \left\{\begin{array}{l} \frac{\partial V_1}{\partial t}+\frac{1}{2} \sigma_1^2 S^2 \frac{\partial^2 V_1}{\partial S^2}+r S \frac{\partial V_1}{\partial S}-r V_1=\lambda_{12}\left(V_1-V_2\right) \\ V_1(0, t)=E e^{-r(T-t)} \\ \lim _{S \rightarrow \infty} V_1(S, t)=0 \\ V_1(S, T)=\max \{E-S, 0\} \end{array}\right. $$
$$ \left\{\begin{array}{l} \frac{\partial V_2}{\partial t}+\frac{1}{2} \sigma_2^2 S^2 \frac{\partial^2 V_2}{\partial S^2}+r S \frac{\partial V_2}{\partial S}-r V_2=\lambda_{21}\left(V_2-V_1\right) \\ V_2(0, t)=E e^{-r(T-t)} \\ \lim _{S \rightarrow \infty} V_2(S, t)=0 \\ V_2(S, T)=\max \{E-S, 0\} \end{array}\right. $$ where $S$ is the value of the underlying asset, $t$ is the current time, $V_j(S, t)(j=1,2)$ is the option value when in state $j$ of the economy, $r$ is the risk-free interest rate (assumed to be constant), $E$ is the strike price and $T$ is the expiration time of the option. We begin by introducing the following dimensionless variables $$ q_j\left(x, \tau_j\right)=\frac{e^x V_j(S, t)}{E}, \quad x=\ln \left(\frac{S}{E}\right), \tau_j=\frac{\sigma_j^2}{2}(T-t) $$ for $j=1$,2. Apart from the inclusion of the exponential factor in $q_j\left(x, \tau_j\right)$, the above change of variables for option valuation problems is a commonly adopted approach to normalize the PDE systems. The exponential factor is included to ensure that the $q_j\left(x, \tau_j\right)$ functions are integrable under the Fourier transform. With the new dimensionless variables, above system become $$ \begin{aligned} & \left\{\begin{array}{l} -\frac{\partial q_1}{\partial \tau_1}+\frac{\partial^2 q_1}{\partial x^2}+\left(\gamma_1-3\right) \frac{\partial q_1}{\partial x}-\left(2 \gamma_1+\beta_{12}-2\right) q_1=-\beta_{12} q_2 \\ \lim _{x \rightarrow-\infty} q_1\left(x, \tau_1\right)=0 \\ \lim _{x \rightarrow \infty} q_1\left(x, \tau_1\right)=0 \\ q_1(x, 0)=\left(e^x-e^{2 x}\right)^{+} \end{array}\right. \\ & \left\{\begin{array}{l} -\frac{\partial q_2}{\partial \tau_2}+\frac{\partial^2 q_2}{\partial x^2}+\left(\gamma_2-3\right) \frac{\partial q_2}{\partial x}-\left(2 \gamma_2+\beta_{21}-2\right) q_2=-\beta_{21} q_1 \\ \lim _{x \rightarrow-\infty} q_2\left(x, \tau_2\right)=0 \\ \lim _{x \rightarrow \infty} q_2\left(x, \tau_2\right)=0 \\ q_2(x, 0)=\left(e^x-e^{2 x}\right)^{+} \end{array}\right. \end{aligned}, $$ where $\gamma_j \equiv \frac{2 r}{\sigma_j^2}$ and $\beta_{j k} \equiv \frac{2 \lambda_{j k}}{\sigma_j^2}$, for $j, k=1,2 j \neq k$, which can be viewed as the interest rate and the rate of leaving each state, relative to the volatility from that state, respectively.
My doubt: I am pretty clear about the fact $$\lim _{x \rightarrow-\infty} q_1\left(x, \tau_1\right)=\lim _{x \rightarrow-\infty} q_2\left(x, \tau_1\right)=0, $$ as $$\lim _{x \rightarrow-\infty} \frac{e^x V_j(S, t)}{E}=0,$$ But how can we see make sure $$\lim _{x \rightarrow \infty} q_1\left(x, \tau_1\right)=\lim _{x \rightarrow \infty} q_2\left(x, \tau_1\right)=0, $$ despite of $$\lim _{S \rightarrow \infty} V_1(S, t)=\lim _{S \rightarrow \infty} V_2(S, t)=0$$ due to the stigma with the factor $\frac{e^x}{E}$?
Your second to last line concerns only limits in the variable x, so given any fixed values for the other variables S, t, E what happens if x goes to negative infinity. The last line is a limit in S. You cannot just switch the limits but as far as I understand it you don't need to.