How do you identify the oblique asymptote of #( 2x^3 + x^2) / (2x^2 - 3x + 3)#?

1 Answer
Tony B
Nov 4, 2015

See explanation.

Explanation:

You have to consider the extreme cases for asymptotes and these will vary according to the equation concerned. Usually it involves #lim_(x-.oo)# but it may also involve#lim_(x->"some number")#. In your case I am assuming we are not looking at the behaviour near #x=0#.

#color(brown)("Consider the numerator:")#
#2x^3# grows much faster than #x^2#. Consequently #2x^3# 'wins' as the dominant factor as #x -> oo#

#color(brown)("Consider the denominator:")#
For the same reason as above, #2x^2# 'wins' as the dominant factor as #x -> oo#.

#color(brown)("Put together:")#
Thus as #x -> oo# we have #lim_(x-> oo) (2x^3+x^2)/(2x^3-3x+3) =(2x^3)/(2x^2) = x#

Note that #x# can be positive or negative.

Related questions
See all questions in Long Division of Polynomials
Impact of this question
1900 views around the world
You can reuse this answer
Creative Commons License