all $t>T_{1}$ and $ x\geq 0$. If $h_{\infty}=+\infty$, then for any given $L$, there exists $\displaystyle{l>\left \{L,\frac{\pi}{2}\sqrt{\frac{d_{1}}{\delta_{1}\delta}}\right\}}$ such that $u$ satisfies $$\label{eq u1}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial u}{\partial t}\geq d_{1}u_{xx}+\delta_{1} u(\delta- u)},&t>T_{1}, 0<x<l, \vspace{7 pt}\\
u_{x}(t,0)= 0, u(t,l)\geq 0,&t>T_{1},\vspace{7 pt}\\
u(T_{1},x)>0, &0<x<l.
\end{array}\right.$$ By Lemma \[lexin3.4\], we know that for sufficiently small $\varepsilon>0,$ $\displaystyle{\liminf_{t\rightarrow+\infty}u(t,x)>\delta-\varepsilon}$ uniformly in any compact subset of $[0,L]$. Since $h_{\infty}=+\infty$, there exists $T_{2}>T_{1}$ such that $h(T_{2})>L$ and $u(t,x)\geq {\delta}/{2} $ for all $t>T_{2}$ and $0\leq x<L$. Then, $(u, v)$ satisfies $$\label{eq vu1}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial u}{\partial t}= d_{1}u_{xx}+u[b_{1}-\delta_{1}(u+v)],}&t>T_{2},~~ 0<x<L, \vspace{7 pt}\\
\displaystyle{\frac{\partial v}{\partial t}=d_{2}v_{xx}+v\left[\frac{b_{2}v}{u+v}-\delta_{2}(u+v)\right],}&t>T_{2},~~ 0<x<L,\vspace{7 pt}\\
u_{x}(t,0)=v_{x}(t,0)=0,&t>T_{2},\vspace{7 pt}\\
\displaystyle{u(T_{2},x)\geq \frac{\delta}{2},} ~~ v(T_{2},x)\leq \kappa_{2}+\delta,&0<x<L.
\end{array}\right.$$ Let $(\underline{u},\bar{v})$ be the solution to the following problem: $$\label{eq vu2}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial\underline{u} }{\partial t}= d_{1}\underline{u}_{xx}+\underline{u}[b_{1}-\delta_{1}(\underline{u}+\bar{v})],}&t>T_{2}, ~~ 0<x<L, \vspace{7 pt}\\

\displaystyle{\frac{\partial\bar{v}}{\partial t}=d_{2}\bar{v}_{xx}+\bar{v}\left[\frac{b_{2}\bar{v}}{\underline{u}+\bar{v}}-\delta_{2}(\underline{u}+\bar{v})\right],}&t>T_{2},~~ 0<x<L,\vspace{7 pt}\\

\partial_{x}\underline{u}(t,0)=\bar{v}_{x}(t,0)=0,&t>T_{2},\vspace{7 pt}\\
\displaystyle{\underline{u}(t,L)= \frac{\delta}{2}}, ~~ \bar{v}(t,L)= \kappa_{2}+\delta,&t>T_{2},\vspace{7 pt}\\
\displaystyle{\underline{u}(T_{2},x)= \frac{\delta}{2},}~~  \bar{v}(T_{2},x)= \kappa_{2}+\delta,&0\leq x\leq L.
\end{array}\right.$$

It follows from the comparison principle that $$u(t,x)\geq \underline{u}(t,x)\text{ and }v(t,x)\leq \bar{v}(t,x)\text{ for }t>T_{2}\text{ and }0\leq x \leq L.$$ By Corollary 3.6 of [@H.SMITH1995], we have $$\displaystyle{\lim_{t\rightarrow+\infty}\underline{u}(t,x)=\underline{u}_{L}(x)}\text{ and }\displaystyle{\lim_{t\rightarrow+\infty}\bar{v}(t,x)=\bar{v}_{L}(x)} \text{ uniformly in }[0,L].$$ Here, $(\underline{u}_{L},\bar{v}_{L})$ satisfies $$\label{eq new-add-a1}
\left\{\begin{array}{ll}
 d_{1}\partial_{xx}\underline{u}_{L}+\underline{u}_{L}[b_{1}-\delta_{1}(\underline{u}_{L}+\bar{v}_{L})]=0,& 0<x<L,\vspace{7 pt} \\
\displaystyle{d_{2}\partial_{xx}\bar{v}_{L}+\bar{v}_{L}\left[\frac{b_{2}\bar{v}_{L}}{\underline{u}_{L}+\bar{v}_{L}}-\delta_{2}(\underline{u}_{L}+\bar{v}_{L})\right]=0,}& 0<x<L,\vspace{7 pt}\\
\partial_{x}\underline{u}_{L}(0)=\partial_{x}\bar{v}_{L}(0)=0,\vspace{7 pt} \\
\displaystyle{\underline{u}_{L}(L)= \frac{\delta}{2}}, ~~ \bar{v}_{L}(L)= \kappa_{2}+\delta.
\end{array}\right.$$ Letting $L\rightarrow+\infty$, 