How do you find the limit of # (x^2 +6) / (6x^3 +x^2 -1)# as x approaches #-oo#?

1 Answer
Andrea S.
Dec 5, 2016

#lim_(x->-oo) frac (x^2+6) (6x^3+x^2-1) = 0#

Explanation:

As the denominator is of higher grade in #x#,

#lim_(x->-oo) frac (x^2+6) (6x^3+x^2-1) = 0#

We can see that by rewriting the expression as such:

#frac (x^2+6) (6x^3+x^2-1) = frac (x^2) (6x^3+x^2-1) + frac 6 (6x^3+x^2-1) = frac 1 (frac (6x^3+x^2-1) (x^2))+ frac 6 (6x^3+x^2-1) = frac 1 (6x+1-1/x^2) + frac 6 (6x^3+x^2-1) #

For both terms of the sum, the numerator is finite, while the denominator grows in absolute value as #x->-oo#, so that the quotient decreases.

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
1141 views around the world
You can reuse this answer
Creative Commons License