Question

Evaluate the following limit??

Answer 1

DNE

Answer 2

;)

Answer 3

(1/t)-(1/t^2+t)

Answer 4

as t->?

Answer 5

t approaches 0

Answer 6

okay

Answer 7

do you just want the answer?

Answer 8

\[\lim_{t\to 0}\frac{1}{t}-\frac{1}{t^2+t}=1\]
why?
\[\frac{1}{t}-\frac{1}{t^2+t}=\frac{1}{t}-\frac{1}{t(t+1)}=\frac{t+1}{t(t+1)}-\frac{1}{t(t+1)}=\frac{t+1-1}{t(t+1)}=\frac{t}{t(t+1)}=\frac{1}{t+1}\]
so
\[\lim_{t\to0}\frac{1}{t}-\frac{1}{t^2+t}=\lim_{t\to0}\frac{1}{t+1}=1\]