How do you evaluate the limit #(2t^2+3t-5)/(1-t)# as t approaches #1#?
1 Answer
Feb 11, 2017
Explanation:
The function is indeterminate for t = 1
#rArrlim_(t to1)(2t^2+3t-5)/(1-t)#
#=lim_(t to1)((2t+5)(t-1))/(1-t)#
#=lim_(t to1)(-(2t+5)cancel((1-t)))/cancel((1-t))# with exclusion t ≠ 1
#rArrlim_(t to1)(-(2t+5))=-7#
