Question

Sol. [Key Concept: Expand $1 - cos(cdots) \approx \frac{1}{2}x^{2}(cdots)$; the limit condition reduces to $\alpha^{2} + 2\alpha - 1 = 0$, so take the product of roots.]

$$\lim_{x \rightarrow 0}\mspace{2mu}\frac{1 - cos\alpha x \cdot cos((\alpha + 1)x)cos((\alpha + 2)x)}{(\alpha + 1)^{2}x^{2}} $$By using L.H. Rule we get

$$\begin{matrix} & \ \Rightarrow \frac{1}{2}(\alpha + 1)^{2}\left\lbrack \alpha^{2} + (\alpha + 1)^{2} + (\alpha + 2)^{2} \right\rbrack = 2 \\ & \alpha^{2} + (\alpha + 1)^{2} + (\alpha + 2)^{2} = 4(\alpha + 1)^{2} \\ & \alpha^{2} + (\alpha + 2)^{2} = 3(\alpha + 1)^{2} \\ & \ \Rightarrow \alpha^{2} + 2\alpha - 1 = 0 \\ & \text{~}\text{Product}\text{~} = - 1 \end{matrix}$$