1. We are asked to find the limit $$\lim_{x \to 4} \frac{3 - \sqrt{5 + x}}{1 - \sqrt{5 - x}}.$$\n\n2. This is a limit involving square roots. Direct substitution gives $x=4$: numerator $3 - \sqrt{9} = 3 - 3 = 0$, denominator $1 - \sqrt{1} = 1 - 1 = 0$. So the limit is of the indeterminate form $\frac{0}{0}$.\n\n3. To resolve this, we use the conjugate to rationalize numerator and denominator. First, multiply numerator and denominator by the conjugate of the numerator: $3 + \sqrt{5 + x}$.\n\n4. Multiply numerator and denominator by $3 + \sqrt{5 + x}$:\n$$\frac{3 - \sqrt{5 + x}}{1 - \sqrt{5 - x}} \cdot \frac{3 + \sqrt{5 + x}}{3 + \sqrt{5 + x}} = \frac{(3)^2 - (\sqrt{5 + x})^2}{(1 - \sqrt{5 - x})(3 + \sqrt{5 + x})} = \frac{9 - (5 + x)}{(1 - \sqrt{5 - x})(3 + \sqrt{5 + x})} = \frac{4 - x}{(1 - \sqrt{5 - x})(3 + \sqrt{5 + x})}.$$\n\n5. Now the limit becomes\n$$\lim_{x \to 4} \frac{4 - x}{(1 - \sqrt{5 - x})(3 + \sqrt{5 + x})}.$$\n\n6. Notice the denominator still has $1 - \sqrt{5 - x}$ which goes to zero as $x \to 4$. We rationalize the denominator by multiplying numerator and denominator by the conjugate of the denominator's first factor: $1 + \sqrt{5 - x}$.\n\n7. Multiply numerator and denominator by $1 + \sqrt{5 - x}$:\n$$\frac{4 - x}{(1 - \sqrt{5 - x})(3 + \sqrt{5 + x})} \cdot \frac{1 + \sqrt{5 - x}}{1 + \sqrt{5 - x}} = \frac{(4 - x)(1 + \sqrt{5 - x})}{(1 - (5 - x))(3 + \sqrt{5 + x})} = \frac{(4 - x)(1 + \sqrt{5 - x})}{(1 - 5 + x)(3 + \sqrt{5 + x})} = \frac{(4 - x)(1 + \sqrt{5 - x})}{(x - 4)(3 + \sqrt{5 + x})}.$$\n\n8. Now we have\n$$\frac{(4 - x)(1 + \sqrt{5 - x})}{(x - 4)(3 + \sqrt{5 + x})} = \frac{\cancel{(4 - x)}(1 + \sqrt{5 - x})}{\cancel{(x - 4)}(3 + \sqrt{5 + x})} = \frac{-(1 + \sqrt{5 - x})}{3 + \sqrt{5 + x}}$$\nbecause $4 - x = -(x - 4)$.\n\n9. Now substitute $x = 4$:\n$$\frac{-(1 + \sqrt{5 - 4})}{3 + \sqrt{5 + 4}} = \frac{-(1 + \sqrt{1})}{3 + \sqrt{9}} = \frac{-(1 + 1)}{3 + 3} = \frac{-2}{6} = -\frac{1}{3}.$$\n\n**Final answer:** $$\boxed{-\frac{1}{3}}.$$