What is lim_(trarr0) (1/t -1/(t^2+t)) ?

2 Answers
Oct 27, 2015

lim_(t->0) (1/t - 1/(t^2+t)) = 1

Explanation:

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

with exclusion t != 0

So

lim_(t->0) (1/t - 1/(t^2+t)) = lim_(t->0) (1/(t+1)) = 1/1 = 1

It is

lim_(trarr0) (1/t -1/(t^2+t))=lim_(trarr0) (t^2+t-t)/(t^2(t+1))=lim_(trarr0) 1/(t+1)=1