What is the limit of #root(3)(x^3-2x^2)-x-1# as #x -> oo# ?

1 Answer
George C.
Feb 27, 2018

#lim_(x->oo) (root(3)(x^3-2x^2)-x-1) = -5/3#

Explanation:

The difference of cubes identity tells us that:

#A^3-B^3 = (A-B)(A^2+AB+B^2)#

Putting #A=(x^3-2x^2)^(1/3)# and #B=(x+1)# we find:

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

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

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

#=-5x^2-3x-1#

So:

#lim_(x->oo) (root(3)(x^3-2x^2)-x-1)#

#=lim_(x->oo) (-(5x^2+3x+1))/((x^3-2x^2)^(2/3)+(x^3-2x^2)^(1/3)(x+1)+(x+1)^2)#

#=lim_(x->oo) (-(5+3/x+1/x^2))/((1-2/x)^(2/3)+(1-2/x)^(1/3)(1+1/x)+(1+1/x)^2)#

#=(-5)/(1+1+1)#

#=-5/3#

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