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`