1. **State the problem:** We want to find the limit $$\lim_{x \to 0} \frac{3 \sin x - \sin 3x}{x^2 \left((2x - 1)^5 + 1\right)}$$ and show that it equals $$\frac{2}{5}$$. 2. **Recall important formulas and rules:** - For small $x$, $\sin x \approx x - \frac{x^3}{6}$ (Taylor expansion). - Binomial expansion for $(1 + u)^n \approx 1 + n u$ when $u$ is small. - Limits involving $\frac{\sin x}{x} \to 1$ as $x \to 0$. 3. **Expand numerator:** Using Taylor expansions, $$3 \sin x = 3 \left(x - \frac{x^3}{6} + O(x^5)\right) = 3x - \frac{3x^3}{6} + O(x^5) = 3x - \frac{x^3}{2} + O(x^5)$$ $$\sin 3x = 3x - \frac{(3x)^3}{6} + O(x^5) = 3x - \frac{27x^3}{6} + O(x^5) = 3x - \frac{9x^3}{2} + O(x^5)$$ So, $$3 \sin x - \sin 3x = \left(3x - \frac{x^3}{2}\right) - \left(3x - \frac{9x^3}{2}\right) + O(x^5) = \left(3x - 3x\right) + \left(-\frac{x^3}{2} + \frac{9x^3}{2}\right) + O(x^5) = 4x^3 + O(x^5)$$ 4. **Expand denominator:** $$(2x - 1)^5 + 1 = \left(-1 + 2x\right)^5 + 1$$ Using binomial expansion, $$(-1 + 2x)^5 = (-1)^5 + 5(-1)^4 (2x) + O(x^2) = -1 + 10x + O(x^2)$$ So, $$ (2x - 1)^5 + 1 = (-1 + 10x + O(x^2)) + 1 = 10x + O(x^2)$$ Therefore, $$x^2 \left((2x - 1)^5 + 1\right) = x^2 (10x + O(x^2)) = 10x^3 + O(x^4)$$ 5. **Form the limit expression:** $$\frac{3 \sin x - \sin 3x}{x^2 \left((2x - 1)^5 + 1\right)} = \frac{4x^3 + O(x^5)}{10x^3 + O(x^4)}$$ 6. **Simplify by canceling $x^3$:** $$= \frac{\cancel{x^3} \left(4 + O(x^2)\right)}{\cancel{x^3} \left(10 + O(x)\right)} = \frac{4 + O(x^2)}{10 + O(x)}$$ 7. **Take the limit as $x \to 0$:** $$\lim_{x \to 0} \frac{4 + O(x^2)}{10 + O(x)} = \frac{4}{10} = \frac{2}{5}$$ **Final answer:** $$\boxed{\frac{2}{5}}$$