## Calculus 10th Edition

$f(x)$ is continuous over the interval $(-\infty, 1)$ U $(1, \infty).$
The only possible point of discontinuity is $x=1\to$ $\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}\dfrac{3x^2-x-2}{x-1}=\lim\limits_{x\to1^-}(3x+2)=3(1^-)+2=5.$ $\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^+}\dfrac{3x^2-x-2}{x-1}=\lim\limits_{x\to1^+}(3x+2)=3(1^+)+2=5.$ Since $\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^-}f(x)\to\lim\limits_{x\to1}f(x)$ exists and is equal to $5.$ But since $\lim\limits_{x\to1}f(x)\ne f(1)$, the function has a discontinuity at $x=1.$ The function is therefore continuous over the interval $(-\infty, 1)$ U $(1, \infty).$