Answer
$f(x)$ is not continuous at $2$.
Work Step by Step
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 $2$.
