If $\lim_{x \to 3} \left( \frac{x^2 - ax - 3a}{x - 3} \right) = 5$, then $a + b =$
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When a limit of a fraction $\frac{f(x)}{g(x)}$ exists finitely at $x=c$ and $g(c)=0$, it is a hard rule that $f(c)$ must also be $0$. Using this initial condition is often enough to solve for unknown parameters without fully evaluating the limit.