#### Answer

We define $f(2)=3$

#### Work Step by Step

Consider the function $f(x)$
$f(x)=\frac{x^2-x-2}{x-2}$
$f(x)=\frac{(x-2)(x+1)}{x-2}$
$f(x)=x+1$
We see that
$\lim\limits_{x\to2}f(x)=\lim\limits_{x\to2}(x+1)=2+1=3$
Since $f(x)$ is continuous at $2$ if and only if $\lim\limits_{x\to2}f(x)=f(2)=3$
Therefore, to make $f(x)$ continuous at $2$, we define $f(2)=3$.