How do you determine the limit of #(x² - 3x - 2)/(x² - 5)# as x approaches infinity?

1 Answer
Johnson Z.
Apr 28, 2016

#lim_(xrarrinfty)(x^2-3x-2)/(x^2-5)=1#

Explanation:

Given,

#lim_(xrarrinfty)(x^2-3x-2)/(x^2-5)#

Divide every term by the term with the highest degree in the denominator.

#=lim_(xrarrinfty)(x^2/color(blue)(x^2)-(3x)/color(blue)(x^2)-2/color(blue)(x^2))/(color(blue)(x^2)/color(blue)(x^2)-5/color(blue)(x^2))#

#=lim_(xrarrinfty)(1-3/x-2/x^2)/(1-5/x^2)#

Substitute #x=infty#.

#=lim_(xrarrinfty)(1-3/infty-2/infty^2)/(1-5/infty^2)#

Any constant divided by #infty# has a limit of #0# since the quotient would approach #0#.

#=lim_(xrarrinfty)(1-0-0)/(1-0)#

#=lim_(xrarrinfty)1/1#

#=1#

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
7059 views around the world
You can reuse this answer
Creative Commons License