If $\lim\limits_{x \to 3}\left(\dfrac{x^2 - ax - 3b}{x - 3}\right) = 5$, then $a + b = $

For the limit to exist and be finite, the expression must take the $\dfrac{0}{0}$ form at $x = 3$.

Substituting $x = 3$ in the numerator gives:

$3^2 - a(3) - 3b = 0$

$9 - 3a - 3b = 0$

$3(a + b) = 9$

$a + b = 3$

Applying L'Hopital's rule to find the individual values:

$\lim\limits_{x \to 3} \dfrac{2x - a}{1} = 5$

$6 - a = 5 \Rightarrow a = 1$

Substituting $a = 1$ into $a + b = 3$ gives $b = 2$.

Thus, $a + b = 1 + 2 = 3$.

Answer: $3$