Answer
$\lim\limits_{t\to0}(\frac{1}{t}-\frac{1}{t^2+t})=1$
Work Step by Step
$\lim\limits_{t\to0}(\frac{1}{t}-\frac{1}{t^2+t})$
$=\lim\limits_{t\to0}(\frac{1}{t}-\frac{1}{t(t+1)})$
$=\lim\limits_{t\to0}\frac{(t+1)-1}{t(t+1)}$
$=\lim\limits_{t\to0}\frac{t}{t(t+1)}$
$=\lim\limits_{t\to0}\frac{1}{t+1}$ (divide both numerator and denominator by $t$)
$=\frac{1}{0+1}$
$=1$
