Answer
$I=\dfrac{V t}{L} $
Work Step by Step
Our aim is to evaluate the limit for $I=\dfrac{V}{R} (1-e^{-Rt/L})$
But $I=\lim\limits_{R \to 0^{+}} \dfrac{V(1-e^{-Rt/L})}{R} =\dfrac{0}{0}$
We can see that the given function shows the indeterminate form of type $\dfrac{0}{0}$. So, we will apply L'Hospital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$I=\lim\limits_{R \to 0^{+}} \dfrac{V(1-e^{-Rt/L})}{R} \\=V \lim\limits_{R \to 0^{+}} \dfrac{e^{-Rt/L}(t/L)}{1}\\=\dfrac{V t}{L} $
