1. **State the problem:** Find the value of $a$ such that the limit $$\lim_{x \to 0} \frac{a \sin x - \sin 2x}{\tan^3 x}$$ is finite. 2. **Recall important formulas and approximations:** - For small $x$, $\sin x \approx x - \frac{x^3}{6}$. - For small $x$, $\tan x \approx x + \frac{x^3}{3}$. - The cube of $\tan x$ for small $x$ is $\tan^3 x \approx x^3$ since higher order terms are negligible for the limit. 3. **Expand numerator using approximations:** - $a \sin x \approx a \left(x - \frac{x^3}{6}\right) = a x - \frac{a x^3}{6}$ - $\sin 2x = 2 \sin x \cos x \approx 2 \left(x - \frac{x^3}{6}\right) \left(1 - \frac{x^2}{2}\right) = 2 \left(x - \frac{x^3}{6} - \frac{x^3}{2} + \text{higher order terms}\right) = 2x - \frac{2x^3}{6} - x^3 = 2x - \frac{x^3}{3} - x^3 = 2x - \frac{4x^3}{3}$ 4. **Form numerator:** $$a \sin x - \sin 2x \approx \left(a x - \frac{a x^3}{6}\right) - \left(2x - \frac{4x^3}{3}\right) = (a - 2) x + \left(-\frac{a}{6} + \frac{4}{3}\right) x^3$$ 5. **Form denominator:** $$\tan^3 x \approx x^3$$ 6. **Form the limit expression:** $$\frac{a \sin x - \sin 2x}{\tan^3 x} \approx \frac{(a - 2) x + \left(-\frac{a}{6} + \frac{4}{3}\right) x^3}{x^3} = \frac{a - 2}{x^2} + \left(-\frac{a}{6} + \frac{4}{3}\right)$$ 7. **Analyze the limit for finiteness:** - The term $\frac{a - 2}{x^2}$ will tend to infinity unless $a - 2 = 0$. - So, set $a - 2 = 0 \Rightarrow a = 2$. 8. **Evaluate the limit with $a=2$:** $$\lim_{x \to 0} \frac{2 \sin x - \sin 2x}{\tan^3 x} = \lim_{x \to 0} \left(-\frac{2}{6} + \frac{4}{3}\right) = -\frac{1}{3} + \frac{4}{3} = \frac{3}{3} = 1$$ **Final answer:** $$a = 2, \quad \lim_{x \to 0} \frac{a \sin x - \sin 2x}{\tan^3 x} = 1$$