# Calculate limmit (8x^3+5x^2)^1/3 -2x when x tend to =infinite?

sciencesolve | Certified Educator

You need to evaluate the following limit such that:

`lim_(x-gtoo) (root(3)(8x^3+5x^2) - 2x) = oo - oo`

The result is indeterminate, hence, you need to multiply `(root(3)(8x^3+5x^2) - 2x)`  by its conjugate to eliminate the cube root such that:

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

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

Notice that you need to use the formula of difference of cubes such that:

`a^3 - b^3 = (a - b)(a^2 + ab + b^2)`

You need to evaluate the limit such that:

`lim_(x->oo) (5x^2)/()(root(3)((8x^3+5x^2)^2) + root(3)(8x^3(8x^3+5x^2)) + root(3)((8x^3)^2)))`

You need to force factor `4x^2`  to denominator such that:

`lim_(x->oo) (5x^2)/(4x^2(root(3)((1+5/(8x))^2) + root(3)((1+5/(8x))) + 1))`

Reducing by `x^2`  and considering the limit `lim_(x->oo) 5/(8x) = 0` , yields:

`lim_(x->oo) (5)/(4(root(3)((1+5/(8x))^2) + root(3)((1+5/(8x))) + 1)) = 5/(4(1+1+1)) = 5/12`

Hence, evaluating the given limit yields `lim_(x->oo) (root(3)(8x^3+5x^2) - 2x) = 5/12.`