Since evaluating the limit yields an indetermination `0/0` , you may also use l'Hospital's theorem, such that:

`lim_(x->3) (x^3 - 27)/(x^2 - 9) = lim_(x->3)((x^3 - 27)')/((x^2 - 9)')`

`lim_(x->3)((x^3 - 27)')/((x^2 - 9)') = lim_(x->3)(3x^2)/(2x)`

Reducing duplicate factors and taking out the constant, yields:

`lim_(x->3)(3x^2)/(2x) = (3/2)lim_(x->3)x = (3/2)*3 = 9/2`

**Hence, evaluating the limit, using l'Hospital's theorem, yields **`lim_(x->3) (x^3 - 27)/(x^2 - 9) = 9/2.`

The limit `lim_(x -> 3) (x^3 - 27)/(x^2 - 9)` has to be determined

`lim_(x -> 3) (x^3 - 27)/(x^2 - 9)`

=> `lim_(x -> 3) ((x-3)*(x^2+3*x+9))/((x - 3)(x + 3))`

=> `lim_(x -> 3) (x^2+3*x+9)/(x + 3)`

Substitute x = 3

=> `(9 + 9 + 9)/(6)`

= `9/2`

**The limit **`lim_(x -> 3) (x^3 - 27)/(x^2 - 9) = 9/2`