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)`

(The entire section contains 2 answers and 132 words.)

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