The limit `lim_(x->6)(x^3 - 216)/(x - 6)` has to be determined.
Substituting x = 6 gives the indeterminate form `0/0` . This allows the use of l'Hopital's rule and the numerator and denominator can be replaced with their derivatives.
=> `lim_(x->6) (3x^2)/1`
Substituting x = 6 gives 108
The limit `lim_(x->6)(x^3 - 216)/(x - 6) = 108`