6. Polynomial

How to take the limit with one polynomial over another when the limit looks like it will be $\dfrac{0}{0}$

Example

$$\lim_{0 \to 3} \dfrac{x^2 - 9}{x^3 - 27}$$

Answer

Factorise

$$x^2 - 9 = (x + 3)(x - 3)$$ $$x^3 - 27 = (x - 3)(x^2 + 3x + 9)$$

since $x \neq 3$

$$\dfrac{x^2 - 9}{x^3 - 27} = \dfrac{x - 3}{x^2 + 3x + 9}$$

therefore

$$\lim_{x \to 3} \dfrac{x - 3}{x^2 + 3x + 9} = \dfrac{3 + 3}{9 + 9 + 9} = \boxed{\dfrac{2}{9}}$$