## Calculus 10th Edition

$f(x)$ is continuous over the interval $(-\infty, \infty).$
$\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}\dfrac{x^2-1}{x-1}=\lim\limits_{x\to1^-}\dfrac{(x+1)(x-1)}{(x-1)}$ $=\lim\limits_{x\to1^-}(x+1)=1^-+1=2.$ $\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^+}\dfrac{x^2-1}{x-1}=\lim\limits_{x\to1^+}\dfrac{(x-1)(x+1)}{(x-1)}$ $=\lim\limits_{x\to1^+}(x+1)=1^++1=2.$ Since $\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^-}f(x)\to\lim\limits_{x\to1}f(x)=2.$ Furthermore from the given, we can conclude that $f(1)=2.$ Since $f(1)=\lim\limits_{x\to1}f(x)\to$ the function is continuous at $x=1$ and hence $f(x)$ is continuous over the interval $(-\infty, \infty).$