How do you find #lim (5t+2)/(t^2-6t+1)# as #t->oo#?
1 Answer
Jan 28, 2017
Explanation:
Divide terms on numerator/denominator by the highest power of t, that is
#t^2#
#rArrlim_(xtooo)(((5t)/t^2+2/t^2)/(t^2/t^2-(6t)/t^2+1/t^2))#
#=lim_(xtooo)((5/t+2/t^2)/(1-6/t+1/t^2))#
#=(0+0)/(1-0+0)=0#
