How do you find the limit of # (8x^2)/(4x^2-3x-1)# as x approaches infinity?
1 Answer
May 29, 2016
2
Explanation:
Divide terms on numerator/denominator by
#x^2#
#((8x^2)/x^2)/((4x^2)/x^2-(3x)/x^2-1/x^2)=8/(4-3/x-1/x^2)# as
#xtooo,3/x" and " 1/x^2to 0#
#rArrlim_(xtooo)(8x^3)/(4x^2-3x-1)=8/4=2#
