## Calculus 10th Edition

$f(x)$ is continuous over the interval $(-\infty, 3)$ U $(3, \infty).$
$\lim\limits_{x\to3^-}f(x)=\lim\limits_{x\to3^-}(2x-4)=2(3^-)-4=2.$ $\lim\limits_{x\to3^+}f(x)=\lim\limits_{x\to3^+}(2x-4)=2(3^+)-4=2.$ Since $\lim\limits_{x\to3^+}f(x)=\lim\limits_{x\to3^-}f(x)\to\lim\limits_{x\to3}f(x)=2.$ From the given, it can be deduced that $f(3)=1.$ Since $\lim\limits_{x\to3}f(x)\ne f(3),$ the function is not continuous at $x=3.$ $f(x)$ is continuous over the interval $(-\infty, 3)$ U $(3, \infty).$