Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to {0^ + }} \left( {\frac{{{e^t}}}{t} - \frac{1}{t}} \right) \cr
& {\text{Combine fractions}} \cr
& = \mathop {\lim }\limits_{t \to {0^ + }} \left( {\frac{{{e^t} - 1}}{t}} \right) \cr
& {\text{evaluating the limit}} \cr
& = \frac{{{e^0} - 1}}{0} = \frac{0}{0} \cr
& {\text{The limit is }}\frac{0}{0}{\text{, so}}{\text{, we can apply l'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{t \to {0^ + }} \frac{{d/dt\left( {{e^t} - 1} \right)}}{{d/dt\left( t \right)}} \cr
& = \mathop {\lim }\limits_{t \to {0^ + }} \frac{{{e^t}}}{1} = \frac{{{e^0}}}{1} = 1 \cr
& {\text{Then}}{\text{,}} \cr
& \mathop {\lim }\limits_{t \to {0^ + }} \left( {\frac{{{e^t}}}{t} - \frac{1}{t}} \right) = 1 \cr} $$
