I have been looking at the following equations in an article* and wanted to know how the $E(∞)$ was derived. By substituting $\gamma E$ for $A$ and factorising it is easy to see how $E(∞)$ can equal 0 but I'm struggling to derive the other answer.
*H. Wang, J. Wang, M. Small, J. M. Moore (2019). Review mechanism promotes knowledge transmission in complex networks. Applied Mathematics and Computation 340, 113-125.
As you noted, at steady-state the third equation leads to
$$ A = \gamma E $$
and using this to eliminate $A$ in the other steady-state equations gives
$$ 0 = (1 - \xi) \gamma E - \lambda \langle k \rangle VE -\delta VE $$
or
$$ 0 = [(1 - \xi) \gamma - \lambda \langle k \rangle V -\delta V] \cdot E $$
This condition can be satisfied when either
$$ E \equiv 0 $$
or
$$ V \equiv \frac{\gamma (1 - \xi)}{\lambda \langle k \rangle + \delta} $$
The article you cite is pay-walled so I haven't read it. Since the model appears to model some population where every member must be either $V$, $E$ or $A$, I assume that these variables represent the fractional portions of the population so that
$$ V(t) + E(t) + A(t) \equiv 1 $$
and also $V$, $E$ and $A$ must be in $[0,1]$.
Under those assumptions, the first steady-state solution is:
$$ E \equiv 0 \implies A \equiv 0 \quad V \equiv 1 $$
To fully specify the second steady-state solution we have
$$ V + E + A \equiv 1 \\ $$
Substituting and solving for $E$ $$ \frac{\gamma (1 - \xi)}{\lambda \langle k \rangle + \delta} + E + \gamma E = 1 \\ E = \frac{1}{1+\gamma} \left\{ 1 - \frac{\gamma (1 - \xi)}{\lambda \langle k \rangle + \delta} \right\} = \frac{1}{1+\gamma} \left\{ 1 - \frac{1}{R_0} \right\} $$
This is the second steady-state from the paper. For $E$ to be non-negative it must be that $R_0 > 1$.