Finding the limit of a fraction: $\lim_{x \to 3} \frac{x^3-27}{x^2-9}$

Wolfram factor the numerator to $(x-3)(x^2+3x+9)$ is there a quick way to find this?

Generally, you have $$ a^{n+1}-b^{n+1}=(a-b)(a^n+a^{n-1}b+a^{n-2}b^2+\cdots+a^2b^{n-2}+ab^{n-1}+b^n) $$ (to see this expand the right hand side, then terms telescope).

Then apply it to $a=x$, $b=3$, $n=2$ giving $$ x^3-27=(x-3)(x^2+3x+9). $$


$$\lim_{x \to 3}\frac{x^3-27}{x^2-9}=\frac{0}{0}=\\\lim_{x \to 3}\frac{(x^3-27)'}{(x^2-9)'} =\\ \lim_{x \to 3}\frac{3x^2}{2x}=\\ \lim_{x \to 3}\frac{3x^2}{2x}\\=\lim_{x \to 3}\frac{3x}{2}=\frac{9}{2}$$

Tags:

Limits

Calculus

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