Answer
$$
\lim _{x \rightarrow 2 } \frac{x^{2}-4}{x-2}=4
$$
Work Step by Step
Given $$
\lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2}
$$
Since from the following table
\begin{array}{|c|c|c|c|c|c|}\hline x\to2^- & {1.9} & {1.99} & {1.999} & {1.9999} \\ \hline \frac{x^{2}-4}{ x-2} & {3.9} & {3.99} & {3.999} & {3.9999} \\ \hline\end{array}
This means that $$
\lim _{x \rightarrow 2^-} \frac{x^{2}-4}{x-2}=4
$$and
\begin{array}{|c|c|c|c|c|}\hline x \to2^+& {2.1} & {2.01} & {2.001} & {2.0001} \\ \hline \frac{x^{2}-4 }{ x-2} & {4.1} & {4.01} & {4.001} & {4.0001} \\ \hline\end{array}
This means that $$
\lim _{x \rightarrow 2^+} \frac{x^{2}-4}{x-2}=4
$$
Hence, this means that $$
\lim _{x \rightarrow 2 } \frac{x^{2}-4}{x-2}=4
$$
