## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$f(x)$ is not continuous at $2$.
When $f(2)=\lim\limits_{x\to 2}f(x)$, then $f(x)$ will be continuous at $x=2$ . $\lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2^-}\dfrac{x^2-2x}{x-2} \\=\lim\limits_{x\to 2^-}\dfrac{x(x-2)}{(x-2)}\\=2$ and $\lim\limits_{x\to 2^+}f(x)=\lim\limits_{x\to 2^+}\dfrac{x-4}{x-1} =-2$ And $f(2)=2$ We can see that our result does not satisfy the statement because $-2 \ne 2$. Therefore, $f(x)$ is not continuous at $2$.