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lim x to infinity (x^3-7x^2)^(1/3)-x

1 Answer
Jim H
Nov 1, 2017

Rewrite using #(a-b)(a^2+ab+b^2) = a^3-b^3#

Explanation:

For the moment, let #m = x^3-7x^2#

#root(3)m-x = ((root(3)m-x))/1((root(3)m^2 + xroot(3)m+x^2))/((root(3)m^2 + xroot(3)m+x^2))#

# =(m-x^3)/(root(3)m^2 + xroot(3)m+x^2)#

# = (x^3-7x^2-x^3)/((root(3)(x^3-7x^2))^2 + xroot(3)(x^3-7x^2)+x^2)#

# = (7x^2)/((xroot(3)(1-7/x))^2+x(xroot(3)(1-7/x))+x^2)#

# = (7x^2)/(x^2((root(3)(1-7/x))^2+root(3)(1-7/x)+1))#

# = (7)/((root(3)(1-7/x))^2+root(3)(1-7/x)+1)#

As #xrarroo#, we have #7/xrarr0#, so the limit is

#7/(1^2+1+1) = 7/3#

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