## Calculus 10th Edition

$f(x)$ is continuous over the interval $(-\infty, 0)$U$(0, \infty).$
Using Theorem $1.12$ with $f(x)=h(g(x))\to h(x)=\cos{x}$ and $g(x)=\dfrac{1}{x}.$ $g(x)$ is continuous for all values of $x$ such that $x\ne0.$ $h(x)$ is continuous for all values of $x.$ Hence, $f(x)=h(g(x))$ is continuous for all values of $x$ such that $x\ne0\to f(x)$ is continuous over the interval $(-\infty, 0)$U$(0, \infty).$