Answer
$$
f(x)=\frac{x^{2}-4}{x-2}
$$
There is a discontinuity when $x=2$.
So
$$
\lim _{x \rightarrow 2} f(x)=2+2=4.
$$
Work Step by Step
$$
f(x)=\frac{x^{2}-4}{x-2}
$$
This rational function is discontinuous wherever the denominator is zero.
There is a discontinuity when $x=2$.
Since for $x \neq 2$
$$
\begin{split}
f(x)&=\frac{x^{2}-4}{x-2} \\
&=\frac{(x-2)(x+2)}{(x-2)} \\
&=(x+2)
\end{split}
$$
So
$$
\lim _{x \rightarrow 2} f(x)=2+2=4.
$$
