We are given,

$\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} – 9}}{{x – 3}}$

$ = \mathop {\lim }\limits_{x \to 3} \frac{{(x – 3)(x + 3)}}{{x – 3}}$

Applying the limits to get,

$ = \mathop {\lim }\limits_{x \to 3} (x + 3) = 6$

Therefore, $\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} – 9}}{{x – 3}} = 6$.