Answer
See the explanation
Work Step by Step
$\begin{array}{lll}
x & \frac{x^2-9}{x-3}\\
2.8&5.8\\
2.9&5.9\\
2.95&5.95\\
2.99&5.99\\
2.999&5.999
\end{array}$
$\begin{array}{lll}
x&\frac{x^2-9}{x-3}\\
3.2&6.2\\
3.1&6.1\\
3.05&6.05\\
3.01&6.01\\
3.001&6.001
\end{array}$
Therefore, $\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3, x\ne3$.
Thus, as $x->3^{-}$, $\frac{x^2-9}{x-3}->6^{-}$ and as $x->3^{+}$, $\frac{x^2-9}{x-3}->6^{+}.$
