Evaluating a limit by combining fractions over a common denominator

Direct substitution at $x=0$ gives $\frac{1}{-3}+\frac{1}{3} = 0$ in the numerator, and dividing by $x=0$ produces the indeterminate form $0/0$. To resolve this, combine the two fractions in the numerator over the common denominator $(x-3)(x+3)$: $\frac{1}{x-3}+\frac{1}{x+3} = \frac{(x+3)+(x-3)}{(x-3)(x+3)} = \frac{2x}{(x-3)(x+3)}.$

Substituting this back into $f(x)$ gives $f(x) = \frac{1}{x}\cdot \frac{2x}{(x-3)(x+3)} = \frac{2x}{x(x-3)(x+3)} = \frac{2}{(x-3)(x+3)},$ after canceling the common factor of $x$ (valid since $x \neq 0$ near the limit point). Now direct substitution works: $\lim_{x\to 0} \frac{2}{(x-3)(x+3)} = \frac{2}{(-3)(3)} = -\frac{2}{9}.$

The common denominator came from multiplying the two original denominators together, namely x minus three and x plus three. This product, x minus three times x plus three, was used to combine the fractions into a single expression. Recognizing this step is what allowed the numerator to simplify to two x, letting the x cancel with the outer denominator.