## University Calculus: Early Transcendentals (3rd Edition)

$f(x)=\frac{x^2+4x-12}{x-2}$ for $x\ne2$ and $f(x)=4$ for $x=2$
$f(x)=\frac{x^2+4x-12}{x-2}$ for $x\ne2$ and $f(x)=4$ for $x=2$ This function is continuous for all $x$ except $x=2$, where it has a removable discontinuity. The reason is stated as follows: 1) For all $x\ne2$, we have $\lim_{x\to c}\frac{x^2+4x-12}{x-2}=\frac{c^2+4c-12}{c-2}=f(c)$, meaning that $f$ is continuous for all $x\ne2$. - For $x=2$, we have $f(2)=4$. Yet, $$\lim_{x\to2}f(x)=\lim_{x\to2}\frac{x^2+4x-12}{x-2}=\lim_{x\to2}\frac{(x-2)(x+6)}{x-2}=\lim_{x\to2}(x+6)$$ $$\lim_{x\to2}f(x)=2+6=8$$ As $\lim_{x\to2}f(x)\ne f(2)$, $f(x)$ is discontinuous at $x=2$. 2) However, this discontinuity is a removable one, because both $\lim_{x\to2}f(x)$ and $f(2)$ still exist; it is just that their values are not equal. If we extend $f(x)$ to include the value $f(2)=\lim_{x\to2}f(x)=8$, then the discontinuity will be removed.