Answer
$-\infty$
Work Step by Step
Consider $f(x)=\lim\limits_{t \to 0^{+}}\dfrac{t-\ln (1+2t)}{t^2}=\lim\limits_{x \to 0} \dfrac{a(x)}{b(x)}$ and $a(0)=0, b(0)=0$
Thus, $f(0)=\dfrac{0}{0}$
This shows an Inderminate form of the limit, so apply L-Hospital's rule:
$\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$
Thus,
$\lim\limits_{t \to 0^{+}}\dfrac{1-2/(1+2t)}{2t}=\dfrac{-1}{0}=-\infty$
