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
So
#lim_(t->0) (1/t - 1/(t^2+t)) = lim_(t->0) (1/(t+1)) = 1/1 = 1#
Oct 27, 2015
It is
