Question

What is the limit of `sqrt[x^3-4x^2+1] + sqrt[x^2-x] +x` as x goes to infinity?

Answer 1

`lim_(x->oo) sqrt[x^3-4x^2+1] + sqrt[x^2-x] +x = oo`

Explanation:

`lim_(x->oo) sqrt[x^3-4x^2+1] = oo`

`lim_(x->oo) sqrt[x^2-x] = oo`

`lim_(x->oo) x = oo`

Since the limit of each term is `oo`, the limit of the sum is `oo`

`lim_(x->oo) sqrt[x^3-4x^2+1] + sqrt[x^2-x] +x = oo`

If you want to give justification for the first 2 limits, use

`sqrt[x^3-4x^2+1] = sqrt(x^3) sqrt(1-4/x+1/x^3)`

and `sqrt[x^2-x] = sqrt(x^2) sqrt(1-1/x)`.

In each case, as `x->oo` the first factor also `->oo` and the second goes to `1`