Calculus: Early Transcendentals 8th Edition

We define $f(2)=3$
Consider the function $f(x)$ $f(x)=\frac{x^3-8}{x^2-4}$ $f(x)=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}$ $f(x)=\frac{x^2+2x+4}{x+2}$ We see that $\lim\limits_{x\to2}f(x)=\lim\limits_{x\to2}\frac{x^2+2x+4}{x+2}=\frac{2^2+2\times2+4}{2+2}=3$ Since $f(x)$ is continuous at $2$ if and only if $\lim\limits_{x\to2}f(x)=f(2)$ Therefore, to make $f(x)$ continuous at $2$, we define $f(2)=3$.