Evaluating a limit by rationalizing the numerator

Direct substitution gives $\frac{\sqrt{6}-\sqrt{6}}{0} = \frac{0}{0}$, an indeterminate form, so we cannot substitute directly. Multiply the numerator and denominator by the conjugate $\sqrt{x+6}+\sqrt{6-x}$: $\frac{\sqrt{x+6}-\sqrt{6-x}}{x}\cdot\frac{\sqrt{x+6}+\sqrt{6-x}}{\sqrt{x+6}+\sqrt{6-x}} = \frac{(x+6)-(6-x)}{x\left(\sqrt{x+6}+\sqrt{6-x}\right)}.$ The numerator simplifies to $2x$.

This leaves $\frac{2x}{x\left(\sqrt{x+6}+\sqrt{6-x}\right)} = \frac{2}{\sqrt{x+6}+\sqrt{6-x}}$ after canceling the common factor of $x$ (valid since $x \ne 0$ as $x \to 0$). Now substituting $x = 0$ directly gives $\frac{2}{\sqrt{6}+\sqrt{6}} = \frac{2}{2\sqrt{6}} = \frac{1}{\sqrt{6}}$, which is the value of the limit.