What is the limit of #(x^(6) + 4x^(4) - 8x^(2)) / [(x^(4) + 12x^(3) + 2x^(2) + 8)]^5# as x goes to infinity?

1 Answer
Jim H
Oct 10, 2015

The limit os #0#

Explanation:

#(x^(6) + 4x^(4) - 8x^(2)) / (x^(4) + 12x^(3) + 2x^(2) + 8)^5# has, as its denominator, a degree 20 polynomial with leading coefficient #1#. Because the degree of the denominator is greater than that of the numerator, we have
#lim(xrarroo)(x^(6) + 4x^(4) - 8x^(2)) / (x^(4) + 12x^(3) + 2x^(2) + 8)^5 =0#

If we want to write some justification, we can use:

#(x^(6) + 4x^(4) - 8x^(2)) / (x^(4) + 12x^(3) + 2x^(2) + 8)^5 = (x^(6) + 4x^(4) - 8x^(2)) / (x^4(1 + 12/x + 2/x^2 + 8/x^4))^5# for #x !=0#

# = (x^20(1/x^14 + 4/x^16 - 8/x^18)) / (x^20(1 + 12/x + 2/x^2 + 8/x^4)^5)#

# = (1/x^14 + 4/x^16 - 8/x^18) / (1 + 12/x + 2/x^2 + 8/x^4)^5#

Now take limits as #xrarroo#, to get

#(0+0+0)/(1+0+0+0)^5 = 0#

Related questions
See all questions in Limits at Infinity and Horizontal Asymptotes
Impact of this question
1466 views around the world
You can reuse this answer
Creative Commons License