Calculus (3rd Edition)

$$\lim _{x \rightarrow 8} \frac{x^2-64}{x-9 }=0$$
Given $$\lim _{x \rightarrow 8} \frac{x^2-64}{x-9}$$ let $$f(x) = \frac{x^2-64}{x-9}$$ Since, we have $$f(8)=\frac{64-64}{8-9}=\frac{0}{-1}=0$$ So, we get \begin{aligned} L&=\lim _{x \rightarrow 9} \frac{x^2-64}{x-9}\\ &= \frac{64-64}{8-9}\\ &= \frac{0}{-1}\\ &=0 \end{aligned}