# Evaluate: `lim_(x-> oo+)sqrt(4x^2 + 5x) - 2x`

### 1 Answer | Add Yours

We have to evaluate: `lim_(x->oo+)sqrt(4x^2 + 5x) - 2x`

If we substitute `x = oo` we get a result of the form `oo - oo` which is indeterminate. But this is not of the form `0/0 or oo/oo` . Therefore l'Hopital's rule cannot be used.

Instead let's change the expression `sqrt(4x^2 + 5x) - 2x`

`sqrt(4x^2 + 5x) - 2x`

=> `((sqrt(4x^2 + 5x) - 2x)(sqrt(4x^2 - 5x) - 2x))/(sqrt(4x^2 + 5x) + 2x)`

=> `((sqrt(4x^2 + 5x))^2 - (2x)^2)/(sqrt(4x^2 + 5x) + 2x)`

=> `(4x^2 + 5x - 4x^2)/(sqrt(4x^2 + 5x) + 2x)`

=> `(5x)/(sqrt(4x^2 + 5x) + 2x)`

=> `(5x)/(x*sqrt(4 + (5/x)) + 2)`

=> `5/(sqrt(4 + (5/x)) + 2)`

`lim_(x->oo+)sqrt(4x^2 + 5x) - 2x`

=> `lim_(x->oo) 5/(sqrt(4 + (5/x)) + 2)`

As `x -> oo, (1/x) -> 0`

Substituting for x we get

`5/(sqrt(4 + 0) + 2)`

=> `5/(2 + 2)`

=> `5/4`

**The required limit is `5/4` **