Answer
We show that $I\left( t \right) \approx \dfrac{{Vt}}{L}$ for small $t$.
Work Step by Step
From Table 2, we know that
${{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
Substituting $ - Rt/L$ for $x$ in the series above gives
${{\rm{e}}^{ - Rt/L}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - Rt/L} \right)}^n}}}{{n!}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{{\left( {Rt/L} \right)}^n}}}{{n!}} = 1 - \dfrac{{Rt}}{L} + \dfrac{{{{\left( {Rt/L} \right)}^2}}}{{2!}} - \dfrac{{{{\left( {Rt/L} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {Rt/L} \right)}^4}}}{{4!}} - \cdot\cdot\cdot$
For small $t$, we get the approximation ${{\rm{e}}^{ - Rt/L}} \approx 1 - \dfrac{{Rt}}{L}$. Therefore,
$I\left( t \right) = \left( {\dfrac{V}{R}} \right)\left( {1 - {{\rm{e}}^{ - Rt/L}}} \right)$
$I\left( t \right) \approx \dfrac{V}{R}\left( {\dfrac{{Rt}}{L}} \right)$
Hence, $I\left( t \right) \approx \dfrac{{Vt}}{L}$ for small $t$.
