1. **Problem:** Find the values of $a$ such that the limit $$\lim_{x \to 0} \frac{\sin 2x + a \sin x}{x^3}$$ exists, and then find the limit. 2. **Formula and rules:** Use Taylor expansions or L'Hôpital's rule for limits of indeterminate forms. Recall: $$\sin x \approx x - \frac{x^3}{6} + \cdots$$ $$\cos x \approx 1 - \frac{x^2}{2} + \cdots$$ 3. **Step 1:** Substitute Taylor expansions: $$\sin 2x \approx 2x - \frac{(2x)^3}{6} = 2x - \frac{8x^3}{6} = 2x - \frac{4x^3}{3}$$ $$a \sin x \approx a \left(x - \frac{x^3}{6}\right) = a x - \frac{a x^3}{6}$$ 4. **Step 2:** Sum numerator: $$\sin 2x + a \sin x \approx (2x + a x) - \left(\frac{4x^3}{3} + \frac{a x^3}{6}\right) = (2 + a) x - \left(\frac{4}{3} + \frac{a}{6}\right) x^3$$ 5. **Step 3:** Divide by $x^3$: $$\frac{\sin 2x + a \sin x}{x^3} \approx \frac{(2 + a) x - \left(\frac{4}{3} + \frac{a}{6}\right) x^3}{x^3} = \frac{2 + a}{x^2} - \left(\frac{4}{3} + \frac{a}{6}\right)$$ 6. **Step 4:** For the limit to exist and be finite as $x \to 0$, the term $\frac{2 + a}{x^2}$ must not blow up, so: $$2 + a = 0 \implies a = -2$$ 7. **Step 5:** Substitute $a = -2$ back: $$\frac{\sin 2x - 2 \sin x}{x^3} \approx - \left(\frac{4}{3} + \frac{-2}{6}\right) = - \left(\frac{4}{3} - \frac{1}{3}\right) = - \frac{3}{3} = -1$$ 8. **Step 6:** Therefore, $$\lim_{x \to 0} \frac{\sin 2x - 2 \sin x}{x^3} = -1$$ **Final answer:** $a = -2$ and the limit is $-1$.