## Calculus (3rd Edition)

We are given the function $f(x)=\tan \left(\frac{1}{x^2+1}\right)$. Since $\frac{1}{x^2+1}$ is a quotient of polynomials, and $1+x^2$ can not be zero for any $x\in R$, then by Theorem 2, the function $\frac{1}{x^2+1}$ is continuous. Now, by Theorem 3, the function $$f(x)=\tan \left(\frac{1}{x^2+1}\right)$$ is continuous.