How do you find the limit #lim (2x^2-3x+2)/(x^3+5x^2-1)# as #x->oo#?

1 Answer
Dec 15, 2016

Answer:

# lim_(x rarr oo) (2x^2-3x+2)/(x^3+5x^2-1) = 0#

Explanation:

Note that as #x rarr oo# then #1/x,1/x^2,1/x^3 rarr 0#

We can therefore multiply numerator and denominator by #1/x^3# (the reciprocal of the largest power in the denominator) as follows:

# lim_(x rarr oo) (2x^2-3x+2)/(x^3+5x^2-1) = lim_(x rarr oo) (2x^2-3x+2)/(x^3+5x^2-1)*(1/x^3)/(1/x^3)#
# " " = lim_(x rarr oo) ((1/x^3)(2x^2-3x+2))/((1/x^3)(x^3+5x^2-1))#
# " " = lim_(x rarr oo) (2/x-3/x^2+2/x^3)/(1+5/x-1/x^3)#
# " " = (0-0+0)/(1+0-0)#
# " " = 0#

We can verify this result by looking at the graph of #y=(2x^2-3x+2)/(x^3+5x^2-1)#
graph{(2x^2-3x+2)/(x^3+5x^2-1) [-7, 13, -4.16, 5.84]}
and indeed it does appear that for large #x# the function is approaching a horizontal asymptote #y=0#