Answer
$\lim_{t\to0}\Big(\dfrac{1}{t}-\dfrac{1}{t^{2}+t}\Big)=1$
Work Step by Step
$\lim_{t\to0}\Big(\dfrac{1}{t}-\dfrac{1}{t^{2}+t}\Big)$
Try to evaluate the limit applying direct substitution:
$\lim_{t\to0}\Big(\dfrac{1}{t}-\dfrac{1}{t^{2}+t}\Big)=\dfrac{1}{0}-\dfrac{1}{0^{2}+0}=\infty-\infty$ Indeterminate form
The limit could not be evaluated using direct substitution. Evaluate the subtraction of fractions and simplify:
$\lim_{t\to0}\Big(\dfrac{1}{t}-\dfrac{1}{t^{2}+t}\Big)=\lim_{t\to0}\Big[\dfrac{(t^{2}+t)-t}{t(t^{2}+t)}\Big]=...$
$...=\lim_{t\to0}\dfrac{t^{2}}{t(t^{2}+t)}=\lim_{t\to0}\dfrac{t^{2}}{t^{2}(t+1)}=\lim_{t\to0}\dfrac{1}{t+1}=...$
Try to evaluate the limit using direct substitution again:
$\lim_{t\to0}\dfrac{1}{t+1}=\dfrac{1}{0+1}=1$
