Answer
Some algebraic manipulation was overseen.
The correct answer is 27.
Work Step by Step
Some algebraic manipulation was overseen.
The numerator is a difference of cubes,
$x^{3}-3^{3}=(x-3)(x^{2}+3x+9),$
so the term can be cancelled out.
The function $g(x)=x^{2}+3x+9$
has the same function values as
$f(x)=\displaystyle \frac{(x-3)(x^{2}+3x+9)}{(x-3)}$
everywhere, except at $x=3$, because $f(3)$ is not defined.
So, we can take:
$\displaystyle \lim_{x\rightarrow 3}f(3)=g(3).$
Thus:
$\displaystyle \lim_{x\rightarrow 3}\frac{(x-3)(x^{2}+3x+9)}{(x-3)}=\lim_{x\rightarrow 3}(x^{2}+3x+9)=9+9+9=27$
