Show that limit of #{(n+1)^(1/3)-n^(1/3)}=0# as x approaches to infinity?

1 Answer
Jim H
Dec 8, 2017

Use the fact that #(a-b)(a^2+ab+b^2) = a^3-b^3#

Explanation:

So, the conjugate of #a^(1/3)-b^(1/3)# is #a^(2/3)+(ab)^(1/3)+b^(2/3)#

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

# = 1/(((x+1)^(2/3)+(x(x+1))^(1/3)+x^(2/3)))#

The limit as #xrarroo# is #1/(oo+oo+oo) = 0#

Related questions
Impact of this question
631 views around the world
You can reuse this answer
Creative Commons License