How do you find #lim (2x^2-4x+1)/(3x^2+5x-6)# as #x->oo#?
1 Answer
Mar 2, 2017
# lim_(x->oo) (2x^2-4x+1)/(3x^2+5x-6) = 2/3 #
Explanation:
We can manipulate the limit as follows
# lim_(x->oo) (2x^2-4x+1)/(3x^2+5x-6) = lim_(x->oo) (2x^2-4x+1)/(3x^2+5x-6) *(1/x^2)/(1/x^2)#
# " "= lim_(x->oo) (2-4/x+1/x^2)/(3+5/x-6/x^2) #
# " "= 2/3 #
As both
