<s>[INST] Finding the limit of a fraction: $\lim_{x \to 3} \frac{x^3-27}{x^2-9}$ 
Find $$\lim_{x \to 3} \frac{x^3-27}{x^2-9}$$

What I did is:
\begin{align}
\lim_{x \to 3} \frac{x^3-27}{x^2-9} &= \lim_{x \to 3} \frac{(x-3)^3+9x-27x}{(x-3)(x+3)} = \lim_{x \to 3} \frac{(x-3)^3+9(x-3)}{(x-3)(x+3)} \\
&= \lim_{x \to 3} \frac{(x-3)^3}{(x-3)(x+3)} + \lim_{x \to 3} \frac{9(x-3)}{(x-3)(x+3)} \\
&= \lim_{x \to 3} \frac{(x-3)^2}{(x+3)} + \lim_{x \to 3} \frac{9}{(x+3)} =0  + \frac{9}{6}
\end{align}
Wolfram factor the numerator to $(x-3)(x^2+3x+9)$ is there a quick way to find this?
 [/INST] $$\lim_{x \to 3}\frac{x^3-27}{x^2-9}=\frac{0}{0}=\\\lim_{x \to 3}\frac{(x^3-27)'}{(x^2-9)'} =\\ \lim_{x \to 3}\frac{3x^2}{2x}=\\ \lim_{x \to 3}\frac{3x^2}{2x}\\=\lim_{x \to 3}\frac{3x}{2}=\frac{9}{2}$$
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