Find the limit?

lim_(nrarr oo)x_n

where

x_n=root(3)((n^2)-(n^3))+n

2 Answers
Mar 23, 2018

1/3

Explanation:

Assuming the limit as

lim_(n->oo)-root(3)(abs(n^2-n^3))+n we have

-root(3)(abs(n^2-n^3))+n = n(-root(3)(abs(1/n-1))+1)=

=(-root(3)(1-1/n)+root(3)(1))/(1/n) = (root(3)(1-1/n)+root(3)(1))/(-1/n)

now calling h = -1/n we have

lim_(n->oo)-root(3)(abs(n^2-n^3))+n = lim_(h->0) (root(3)(1+h)+root(3)(1))/h = 1/3

Mar 23, 2018

Please see below.

Explanation:

Use (a+b)(a^2-ab+b^2) = a^3+b^3 to remove the third root in the numerator.

root(3)(n^2-n^3) + n = ((root(3)(n^2-n^3) + n ))/1 * (((root(3)(n^2-n^3))^2 - nroot(3)(n^2-n^3)+n^2)) / (((root(3)(n^2-n^3))^2 - nroot(3)(n^2-n^3)+n^2))

= (n^2-n^3+n^3)/ ((root(3)(n^2-n^3))^2 - nroot(3)(n^2-n^3)+n^2)

= n^2/ (n^2(root(3)(1/n-1))^2 - n^2root(3)(1/n-1)+n^2)

= 1/ ((root(3)(1/n-1))^2 - root(3)(1/n-1)+1)

So,

lim_(nrarroo)(root(3)(n^2-n^3) + n ) = lim_(nrarroo) 1/ ((root(3)(1/n-1))^2 - root(3)(1/n-1)+1)

= 1/ ((root(3)(0-1))^2 - root(3)(0-1)+1)

= 1/ ((-1)^2 - (-1)+1)

= 1/3