Evaluating a limit by combining fractions over a common denominator

Combine the terms in the numerator over a common denominator: $\frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x}$. So the expression becomes $s(x) = \frac{\frac{3-x}{3x}}{x-3} = \frac{3-x}{3x(x-3)}.$

Note that $3 - x = -(x-3)$, so we can cancel the common factor $(x-3)$ (valid since $x \ne 3$ as $x \to 3$): $s(x) = \frac{-(x-3)}{3x(x-3)} = \frac{-1}{3x}.$ Now direct substitution is valid since $\frac{-1}{3x}$ is continuous at $x=3$: $\lim_{x\to 3} \frac{-1}{3x} = \frac{-1}{9}.$

After canceling the common factor, the expression simplified to negative one divided by three x. This simplified form is continuous at x equals three, so direct substitution could finally be applied. Substituting x equals three into this simplified expression gives negative one ninth, matching the final answer from part a.