$$\lim_{x\to 0}\bigg( \Big( \frac{1+\sin(x)\cos(\alpha x)}{1+\sin(x)\cos(\beta x)} \Big) ^{\cot^3(x)}\bigg)$$ $$\lim_{x\to 0} \bigg( \Big( \frac{1+\sin(x)\cos(\alpha x)+\sin(x)\cos(\beta x)-\sin(x)\cos(\beta x)}{1+\sin(x)\cos(\beta x)} \Big) ^{\cot^3(x)}\bigg)$$
$$\lim_{x\to 0} \bigg( \Big( 1+\frac{\sin(x)\cos(\alpha x)-\sin(x)\cos(\beta x)}{1+\sin(x)\cos(\beta x)}\Big)^{\cot^3(x) \cdot \frac{1+\sin(x)\cos(\beta x)}{\sin(x)\cos(\alpha x)-\sin(x)\cos(\beta x)} \cdot \frac{\sin(x)\cos(\alpha x)-\sin(x)\cos(\beta x)}{1+\sin(x)\cos(\beta x)}} \bigg) $$
$$e^{\lim_{x\to 0} \big( \frac{\cos^3(x)\sin(x)(\cos(\alpha x) - \cos(\beta x))}{\sin(x)\sin^2(x) (1+\sin(x)\cos(\beta x))} \big)}$$
Currently disregarding the exponent, because the font is small.
$$\lim_{x\to 0} \big( \frac{\cos(\alpha x) - \cos(\beta x)}{\sin^2(x)} \big)$$
Through Taylor's expansion,
$$\lim_{x\to 0} \big( \frac{1-\frac{\alpha^2x^2}{2} -1 + \frac{\beta^2x^2}{2}}{x^2} \big)$$
And the answer is
$$e^{\frac{\beta^2 - \alpha^2}{2}}.$$
The answer in the book, however, is $e^{\beta^2 - \alpha^2}$. The formula for the difference of two cosines also leaves me with $e^{\frac{\beta^2 - \alpha^2}{2}}$, and I feel like there's a typo, but maybe I went wrong somewhere?
There's a typo, as affirmed in the comments.