## Calculus: Early Transcendentals 8th Edition

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