1. **State the problem:** Find the limit $$\lim_{x \to 4} \frac{6 - \sqrt{x + 5} \cdot \sqrt[3]{x + 4}}{2\sqrt{x + 5} - 3\sqrt[3]{x + 4}}$$ 2. **Evaluate direct substitution:** Substitute $x=4$: - $\sqrt{4+5} = \sqrt{9} = 3$ - $\sqrt[3]{4+4} = \sqrt[3]{8} = 2$ Numerator: $$6 - (3)(2) = 6 - 6 = 0$$ Denominator: $$2(3) - 3(2) = 6 - 6 = 0$$ We get an indeterminate form $\frac{0}{0}$, so we need to simplify. 3. **Rewrite expressions:** Let $$a = \sqrt{x + 5} = (x+5)^{1/2}, \quad b = \sqrt[3]{x + 4} = (x+4)^{1/3}$$ Our expression becomes: $$\frac{6 - a b}{2a - 3b}$$ 4. **Apply substitution near $x=4$:** Define $$h = x - 4, \quad \text{so} \quad x=4+h, h \to 0$$ Then $$a = (9+h)^{1/2} \approx 3 + \frac{h}{2 \cdot 3} = 3 + \frac{h}{6}$$ $$b = (8+h)^{1/3} \approx 2 + \frac{h}{3 \cdot 2^{2}} = 2 + \frac{h}{12}$$ (using linear approximation: $(c+h)^r \approx c^r + r c^{r-1} h$) 5. **Substitute approximations into numerator and denominator:** Numerator: $$6 - a b \approx 6 - \left(3 + \frac{h}{6}\right) \left(2 + \frac{h}{12}\right)$$ Multiply inside: $$= 6 - \left(3 \cdot 2 + 3 \cdot \frac{h}{12} + 2 \cdot \frac{h}{6} + \frac{h}{6} \cdot \frac{h}{12}\right)$$ $$= 6 - \left(6 + \frac{3h}{12} + \frac{2h}{6} + \text{higher order } h^2 \text{ terms} \right)$$ Simplify coefficients: $$\frac{3h}{12} = \frac{h}{4}, \quad \frac{2h}{6} = \frac{h}{3}$$ So numerator: $$6 - \left(6 + \frac{h}{4} + \frac{h}{3} + O(h^2) \right) = 6 - 6 - \frac{h}{4} - \frac{h}{3} - O(h^2) = - \left( \frac{h}{4} + \frac{h}{3} \right) + O(h^2)$$ $$= - \left( \frac{3h}{12} + \frac{4h}{12} \right) + O(h^2) = - \frac{7h}{12} + O(h^2)$$ Denominator: $$2a - 3b \approx 2 \left(3 + \frac{h}{6}\right) - 3 \left(2 + \frac{h}{12}\right) = 6 + \frac{2h}{6} - 6 - \frac{3h}{12}$$ Simplify coefficients: $$\frac{2h}{6} = \frac{h}{3}, \quad \frac{3h}{12} = \frac{h}{4}$$ So denominator: $$6 + \frac{h}{3} - 6 - \frac{h}{4} = \frac{h}{3} - \frac{h}{4} = \frac{4h}{12} - \frac{3h}{12} = \frac{h}{12}$$ 6. **Form the limit from approximations:** $$\lim_{h \to 0} \frac{- \frac{7h}{12} + O(h^2)}{\frac{h}{12} + O(h^2)} = \lim_{h \to 0} \frac{-7h/12}{h/12} = -7$$ 7. **Final answer:** $$\boxed{-7}$$