How do you find the limit of #(8x^3 + 5x^2)^(1/3) -2x # as x approaches infinity?

1 Answer
Oct 21, 2016

Answer:

Rationalize and simplify.

Explanation:

Recall that #a^3-b^3 = (a-b)(a^2+ab+b^2)#.

This allows us to rationalize #root(3)u - v# by multiplying by

#(root(3)u)^2+vroot(3)u+v^2# to get #u-v^3#.

We'll do that with #u = root(3)(8x^3+5x^2)# and #v = 2x#.

#((root(3)(8x^3 + 5x^2) -2x))/1 * (((root(3)(8x^3 + 5x^2))^2+2x root(3)(8x^3 + 5x^2) + (2x)^2))/ (((root(3)(8x^3 + 5x^2))^2+2x root(3)(8x^3 + 5x^2) + (2x)^2))#

# = ((root(3)(8x^2+5x^2))^3 - (2x)^3)/ (((root(3)(8x^3 + 5x^2))^2+2x root(3)(8x^3 + 5x^2) + (2x)^2))#

# = (8x^2+5x^2 - 8x^3)/ (((root(3)(8x^3 + 5x^2))^2+2x root(3)(8x^3 + 5x^2) + 4x^2))#

# = (5x^2)/ (((root(3)(8x^3 + 5x^2))^2+2x root(3)(8x^3 + 5x^2) + 4x^2))#

Now we'll do some work in the denominator to factor out an #x^2#.

For #x != 0#, we have

# = (5x^2)/ (((root(3)(x^3(8 + 5/x)))^2+2x root(3)(x^3(8 + 5/x) + 4x^2))#

# = (5x^2)/ (((xroot(3)((8 + 5/x)))^2+2x^2 root(3)(8 + 5/x) + 4x^2))#

# = (5x^2)/ (x^2((root(3)((8 + 5/x)))^2+2 root(3)(8 + 5/x) + 4))#

# = 5/ ((root(3)((8 + 5/x)))^2+2 root(3)(8 + 5/x) + 4)#.

Evaluating the limit as #xrarroo#, we finish with

# = 5/((root(3)((8 + 0)))^2+2 root(3)(8 + 0) + 4)#

# = 5/(4+4+4) = 5/12#.