The value of `lim_(x->oo)(6x^3+x-6)/(-4x^3+9)` has to be determined.

It is seen that substituting x = `oo` gives the result `oo/oo` which is indeterminate. This allows the use of l'Hopital's rule and the numerator and denominator can be replaced by their derivatives.

=> `lim_(x->oo)(18x^2 + 1)/(-12x^2)`

Again, substituting x = `oo` gives an the indeterminate form` oo/oo` . Replace the numerator and denominator by their derivatives.

=> `lim_(x->oo) (36x)/(-24x)`

Again, replace the numerator and denominator by their derivatives

=> `lim_(x->oo) -36/24`

=> `-3/2`

**The value of **`lim_(x->oo)(6x^3+x-6)/(-4x^3+9) = -3/2`

The limit `lim_(x->oo) (6x^3 + x - 6)/(-4x^3 + 9x)` has to be determined.

If we substitute `x = oo` in `(6x^3 + x - 6)/(-4x^3 + 9x)` , the result is `oo/oo` which is indeterminate.

To determine the limit we can rewrite the expression as follows:

`lim_(x->oo) (6x^3 + x - 6)/(-4x^3 + 9x)`

`= lim_(x->oo) ((6x^3 + x - 6)/x^3)/((-4x^3 + 9x)/x^3)`

`= lim_(x->oo) (6 + 1/x^2 - 6/x^3)/(-4 + 9/x^2)`

As `x->oo` , `1/x -> 0` and `1/x^2` and `1/x^3` also tend to 0

Substituting `x = oo` in the expression gives:

`(6 + 1/(oo)^2 - 6/(oo)^3)/(-4 + 9/(oo)^2)`

= `(6 + 0 - 0)/(-4 + 0)`

= `-3/2`

The required limit is `-3/2` .