Find the limit?

#lim_(nrarr oo)x_n#

where

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

2 Answers
Cesareo R.
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#

Jim H
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#

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