Evaluate a limit by combining fractions over a common denominator

First, combine the numerator $\frac{1}{x+3} - \frac{1}{3}$ over the common denominator $3(x+3)$: $\frac{1}{x+3} - \frac{1}{3} = \frac{3 - (x+3)}{3(x+3)} = \frac{-x}{3(x+3)}.$ This rewrites the original expression as $\frac{-x}{3(x+3)} \cdot \frac{1}{x} = \frac{-x}{3x(x+3)}.$

Since $x \to 0$ but $x \neq 0$, we can cancel the common factor of $x$ from numerator and denominator, giving $\frac{-1}{3(x+3)}.$ This function is continuous at $x=0$, so direct substitution now gives $\frac{-1}{3(0+3)} = \frac{-1}{9}.$