Answer
$$ - \infty $$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to {0^ + }} \frac{{t - \ln \left( {1 + 2t} \right)}}{{{t^2}}} \cr
& {\text{Evaluating the limit gives}} \cr
& \mathop {\lim }\limits_{t \to {0^ + }} \frac{{t - \ln \left( {1 + 2t} \right)}}{{{t^2}}} = \frac{{0 - \ln \left( {1 + 0} \right)}}{{{0^2}}} = \frac{0}{0} \cr
& {\text{The limit is }}\frac{0}{0}{\text{; so}}{\text{, we can apply l'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{t \to {0^ + }} \frac{{d/dt\left( {t - \ln \left( {1 + 2t} \right)} \right)}}{{d/dt\left( {{t^2}} \right)}} \cr
& = \mathop {\lim }\limits_{t \to {0^ + }} \frac{{1 - \frac{2}{{1 + 2t}}}}{{2t}} \cr
& = \mathop {\lim }\limits_{t \to {0^ + }} \frac{{1 - \frac{2}{{1 + 2t}}}}{{2t}} \cr
& {\text{evaluating the limit}} \cr
& = \frac{{1 - \frac{2}{{1 + 2\left( 0 \right)}}}}{{2\left( 0 \right)}} \cr
& = \frac{{ - 1}}{0} = \pm \infty \cr
& {\text{The limit tends to }}{{\text{0}}^ + }{\text{. Then}}{\text{,}} \cr
& = \frac{{ - 1}}{{{0^ + }}} = - \infty \cr
& \cr
& \mathop {\lim }\limits_{t \to {0^ + }} \frac{{t - \ln \left( {1 + 2t} \right)}}{{{t^2}}} = - \infty \cr} $$
