Question

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

Answer 1

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`.