Question:

Put $x = 1 + h$, so as $x \to 1^+$, we have $h \to 0^+$. $$\lim_{h \to 0^+} \frac{h(6 + \lambda \cos h) - \mu \sin h}{h^3} = -1$$ Using Taylor series expansions: - $\cos h = 1 - \frac{h^2}{2!} + \frac{h^4}{4!} - \cdots$ - $\sin h = h - \frac{h^3}{3!} + \cdots$ $$\lim_{h \to 0^+} \frac{h\left(6 + \lambda\left(1 - \frac{h^2}{2} + \cdots\right)\right) - \mu\left(h - \frac{h^3}{6} + \cdots\right)}{h^3} = -1$$ $$\lim_{h \to 0^+} \frac{6h + \lambda h - \frac{\lambda h^3}{2} - \mu h + \frac{\mu h^3}{6} + \cdots}{h^3} = -1$$ For the limit to exist and equal $-1$, the coefficient of $h$ must be zero: $$6 + \lambda - \mu = 0$$ The coefficient of $h^3$ gives: $$-\frac{\lambda}{2} + \frac{\mu}{6} = -1$$ From the first equation: $\mu = 6 + \lambda$ Substituting into the second equation: $$-\frac{\lambda}{2} + \frac{6 + \lambda}{6} = -1$$ $$-\frac{3\lambda}{6} + \frac{6 + \lambda}{6} = -1$$ $$\frac{-3\lambda + 6 + \lambda}{6} = -1$$ $$-2\lambda + 6 = -6$$ $$\lambda = 6$$ Therefore: $\mu = 6 + 6 = 12$ $$\lambda + \mu = 6 + 12 = 18$$