Answer
$$ - 1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to 0} \frac{{t{e^t}}}{{1 - {e^t}}} \cr
& {\text{Evaluate using theorem 1}}{\text{.2}}{\text{.2 }}\left( {{\text{law }}\left( d \right)} \right) \cr
& = \frac{{\mathop {\lim }\limits_{t \to 0} t{e^t}}}{{\mathop {\lim }\limits_{t \to 0} \left( {1 - {e^t}} \right)}} \cr
& {\text{Using law }}\left( a \right) \cr
& = \frac{{\mathop {\lim }\limits_{t \to 0} t{e^t}}}{{\mathop {\lim }\limits_{t \to 0} 1 - \mathop {\lim }\limits_{t \to 0} {e^t}}} \cr
& = \frac{{\left( 0 \right){e^0}}}{{1 - {e^0}}} = \frac{0}{{1 - 1}} \cr
& = \frac{0}{0} \cr
& {\text{The numerator and denominator have a limit of 0}}{\text{, so the limit is }} \cr
& {\text{an indeterminate form of type 0/0}}{\text{. Use L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{t \to 0} \frac{{t{e^t}}}{{1 - {e^t}}} = \mathop {\lim }\limits_{t \to 0} \frac{{d/dt\left( {t{e^t}} \right)}}{{d/dt\left( {1 - {e^t}} \right)}} \cr
& = \mathop {\lim }\limits_{t \to 0} \frac{{t{e^t} + {e^t}}}{{ - {e^t}}} \cr
& {\text{evaluating}} \cr
& = \frac{{\left( 0 \right){e^0} + {e^0}}}{{ - {e^0}}} \cr
& = \frac{{0 + 1}}{{ - 1}} = - 1 \cr
& {\text{then }}\mathop {\lim }\limits_{t \to 0} \frac{{t{e^t}}}{{1 - {e^t}}} = - 1 \cr} $$
