1. **State the problem:** Find the limit $$\lim_{x \to -3} \frac{ \sqrt{2}\sqrt{x^2 - x} - \sqrt{6}\sqrt{x^2 + x - 6} }{ x^2 - 4x + 3 }$$ 2. **Analyze the expression:** The denominator factors as $$x^2 - 4x + 3 = (x - 3)(x - 1)$$ 3. **Check direct substitution:** Substitute $x = -3$: - Numerator: $\sqrt{2}\sqrt{9 + 3} - \sqrt{6}\sqrt{9 - 3 - 6} = \sqrt{2}\sqrt{12} - \sqrt{6}\sqrt{0} = \sqrt{2} \times 2\sqrt{3} - 0 = 2\sqrt{6}$ - Denominator: $9 + 12 + 3 = 24$ (incorrect, re-check denominator) Actually, denominator at $x=-3$: $$(-3)^2 - 4(-3) + 3 = 9 + 12 + 3 = 24$$ So denominator is 24, numerator is $2\sqrt{6}$, so limit is $\frac{2\sqrt{6}}{24} = \frac{\sqrt{6}}{12}$. 4. **Conclusion:** Since direct substitution does not give an indeterminate form, the limit is simply $$\boxed{\frac{\sqrt{6}}{12}}$$