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 ** ...

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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`