Answer
See below
Work Step by Step
The different equation of the RC circuit is:
$\frac{dq}{dt}+\frac{q}{RC}=\frac{E_0}{R}$
The integrating factor is:
$I(t)=e^{\int\frac{1}{RC} dt}=e^{t/RC}$
Multiply both sides by integrating factor:
$e^{t/RC}\frac{dq}{dt}+e^{t/RC}\frac{q}{RC}=e^{t/RC}\frac{E_0}{R}\\
\frac{d}{dt}(qe^{t/RC})=e^{t/RC}\frac{E_0}{R}$
Integrating both sides, we have:
$qe^{t/RC}=c_1+\int e^{t/RC}\frac{E_0}{R}dt\\qe^{t/RC}=E_0Ce^{t/RC}\\
\rightarrow q(t)=c_1e^{-t/RC}+E_0C$
Since $q(0)=0$, $q(0)=c_1+E_0C=0\\
\rightarrow c_1=-E_0C$
Therefore, $q(t)=-E_0Ce^{-t/RC}+E_0C$
With $t \rightarrow \infty$, then $\lim(-E_0Ce^{-t/RC}+E_0C)=E_0C$
The current in the circuit:
$i(t)=\frac{dq}{dt}=\frac{E_0C}{RC}e^{-t/RC}$
With $t \rightarrow \infty$, then $\lim i(t)=0$
