The current $\left(i\right)$  at time $t=0$  and  $t=\infty$ respectively for the given circuit is:

$\begin{array}{l}\text{ At }\mathrm{t}=0,\text{ Inductor behaves as infinite resistance. }\\ \text{ So, }{\mathrm{R}}_{\text{net }}=(5+1)\mathrm{and}(5+4) \mathrm{are} \mathrm{in} \mathrm{parallel} \mathrm{combination}\\ {\mathrm{R}}_{\mathrm{net}}=\frac{6\times 9}{6+9}=\frac{54}{15}=\frac{18}{5}\mathrm{\Omega }\\ \text{ Thus, }\mathrm{i}=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{net}}}=\frac{5\mathrm{E}}{18}\end{array}$

$\begin{array}{l}\text{ At }t=\mathrm{\infty }\text{ , Inductor behaves as conducting wire. }\\ \text{ So, }{R}_{\text{net }}=\left(5\right|\left|5\right)+\left(1\right|\left|4\right)=\frac{5}{2}+\frac{4}{5}=\frac{33}{10}\mathrm{\Omega }\\ \text{ Thus, }i=\frac{E}{R_{\text{net }}}=\frac{10E}{33}\end{array}$