1. Stating the problem: Find the value of $$16(a^2 + b^2 + c^2)$$ given that $$\lim_{x \to 0} \frac{ax^2 e^x - b \ln(1+x) + cx e^{-x}}{x^2 \sin x} = 1$$ and this limit is finite. 2. Analyzing the limit: Recall that as $$x \to 0$$, $$e^x = 1 + x + \frac{x^2}{2} + O(x^3),$$ $$e^{-x} = 1 - x + \frac{x^2}{2} + O(x^3),$$ $$\ln(1+x) = x - \frac{x^2}{2} + O(x^3),$$ $$\sin x = x - \frac{x^3}{6} + O(x^5).$$ 3. Expand each term in the numerator: $$ax^2 e^x = a x^2 \left(1 + x + \frac{x^2}{2}\right) = a x^2 + a x^3 + \frac{a x^4}{2} + O(x^5),$$ $$-b \ln(1+x) = -b \left(x - \frac{x^2}{2}\right) = -b x + \frac{b x^2}{2} + O(x^3),$$ $$c x e^{-x} = c x \left(1 - x + \frac{x^2}{2}\right) = c x - c x^2 + \frac{c x^3}{2} + O(x^4).$$ 4. Summing the numerator terms: $$a x^2 + a x^3 + \frac{a x^4}{2} - b x + \frac{b x^2}{2} + c x - c x^2 + \frac{c x^3}{2} + O(x^3).$$ Group by powers of $$x$$: $$(-b + c) x + \left(a + \frac{b}{2} - c\right) x^2 + \left(a + \frac{c}{2}\right) x^3 + O(x^4).$$ 5. Expand denominator: $$x^2 \sin x = x^2 \left(x - \frac{x^3}{6}\right) = x^3 - \frac{x^5}{6} + O(x^7).$$ 6. Form the quotient: $$\frac{\text{numerator}}{\text{denominator}} = \frac{(-b + c) x + \left(a + \frac{b}{2} - c\right) x^2 + \left(a + \frac{c}{2}\right) x^3 + O(x^4)}{x^3 + O(x^5)}.$$ 7. For the limit to be finite as $$x \to 0$$, the terms with lower powers than $$x^3$$ in numerator must vanish: - Coefficient of $$x$$ must be zero: $$-b + c = 0 \implies c = b.$$ - Coefficient of $$x^2$$ must be zero: $$a + \frac{b}{2} - c = 0.$$ Substitute $$c = b$$ gives: $$a + \frac{b}{2} - b = a - \frac{b}{2} =0 \implies a = \frac{b}{2}.$$ 8. Now numerator reduces to: $$\left(a + \frac{c}{2}\right) x^3 + O(x^4) = \left( \frac{b}{2} + \frac{b}{2} \right) x^3 + O(x^4) = b x^3 + O(x^4).$$ 9. The expression becomes: $$\frac{b x^3 + O(x^4)}{x^3 + O(x^5)} \to b$$ as $$x \to 0$$. Given the limit equals 1, we have $$b = 1.$$ 10. Using $$b =1$$ find $$a$$ and $$c$$: $$a = \frac{1}{2}, \quad c = 1.$$ 11. Compute $$16(a^2 + b^2 + c^2)$$: $$16 \left( \left(\frac{1}{2}\right)^2 + 1^2 + 1^2 \right) = 16 \left( \frac{1}{4} + 1 + 1 \right) = 16 \cdot \frac{9}{4} = 36.$$ Check options: none matches 36 exactly, so re-check step 10. 12. Re-examine step 10: We found $$a = \frac{b}{2} = \frac{1}{2}$$ and $$c = b =1$$ Then sum of squares: $$a^2 + b^2 + c^2 = \left(\frac{1}{2}\right)^2 + 1^2 + 1^2 = \frac{1}{4} + 1 + 1 = \frac{9}{4} = 2.25.$$ Hence, $$16 \times 2.25 = 36,$$ which is not one of the options. 13. Check if denominator expansion used correctly: Recall $$\sin x = x - \frac{x^3}{6} + \dots$$ so $$x^2 \sin x = x^3 - \frac{x^5}{6} + O(x^7)$$ so dominant term is $$x^3$$. 14. Limit equals coefficient of $$x^3$$ in numerator over 1. From numerator coefficient of $$x^3$$: $$a + \frac{c}{2} = 1$$ (must equal 1 since limit is 1). Substitute $$a = \frac{b}{2}$$ and $$c = b$$ gives: $$\frac{b}{2} + \frac{b}{2} = b = 1,$$ which matches step 9. 15. Confirm the numerator coefficient of $$x^3$$ is indeed $$a + \frac{c}{2}$$ using the expansions. From step 3: - $ax^2 e^x$ term: $$a x^3$$ - $c x e^{-x}$ term: $$\frac{c x^3}{2}$$ Sum: $$a x^3 + \frac{c x^3}{2}$$, which is as used. 16. Therefore, $$a = \frac{1}{2}, b = 1, c = 1$$ and $$16 (a^2 + b^2 + c^2) = 16 \left( \frac{1}{4} + 1 + 1 \right) = 16 \cdot \frac{9}{4} = 36.$$ 17. Since 36 is not listed, check simplification carefully for the $$b \ln(1+x)$$ term sign and expansion again. Original term: $$-b \ln(1+x)$$ expands to $$-b \left(x - \frac{x^2}{2} + \frac{x^3}{3} \right) = -b x + \frac{b x^2}{2} - \frac{b x^3}{3}.$$ 18. Now, cumulative numerator up to $$x^3$$: $$a x^2 + a x^3 - b x + \frac{b x^2}{2} - \frac{b x^3}{3} + c x - c x^2 + \frac{c x^3}{2}.$$ Group by powers: - $$x$$: $$-b + c,$$ - $$x^2$$: $$a + \frac{b}{2} - c,$$ - $$x^3$$: $$a - \frac{b}{3} + \frac{c}{2}.$$ 19. Conditions for finite limit: $$-b + c = 0 \Rightarrow c = b$$ $$a + \frac{b}{2} - c = 0 \Rightarrow a + \frac{b}{2} - b = 0 \Rightarrow a = \frac{b}{2}$$ 20. Limit equals coefficient of $$x^3$$ in numerator over $$x^3$$ denominator leading term 1, so $$a - \frac{b}{3} + \frac{c}{2} = 1.$$ Substitute $$a = \frac{b}{2}, c = b$$: $$\frac{b}{2} - \frac{b}{3} + \frac{b}{2} = 1 \implies \left(\frac{b}{2} + \frac{b}{2} - \frac{b}{3}\right) = 1 \implies \left(b - \frac{b}{3}\right) = 1 \implies \frac{2b}{3} = 1 \implies b = \frac{3}{2}.$$ 21. Then: $$c = b = \frac{3}{2}, \quad a = \frac{b}{2} = \frac{3}{4}.$$ 22. Calculate $$16(a^2 + b^2 + c^2)$$: $$16 \left( \left(\frac{3}{4}\right)^2 + \left(\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right)^2 \right) = 16 \left( \frac{9}{16} + \frac{9}{4} + \frac{9}{4} \right) = 16 \left( \frac{9}{16} + \frac{18}{4} \right).$$ Convert $$\frac{18}{4} = \frac{72}{16}$$ to common denominator: $$= 16 \cdot \left( \frac{9}{16} + \frac{72}{16} \right) = 16 \cdot \frac{81}{16} = 81.$$ 23. Final answer: $$\boxed{81}.$$