1. **Stating the problem:** Calculate the limit $$\lim_{x \to 0} \frac{e^{2x} \sin(x^2) + 2 \cos x - 2 e^{-x^3}}{(\ln(2+x) - \ln 2) \sin(x^2 e^{-x}) \arctan(x + \frac{1}{x})}$$ 2. **Analyze the expression:** We need to evaluate the behavior of numerator and denominator as $x \to 0$. 3. **Approximate each component near 0 using Taylor expansions:** - $e^{2x} \approx 1 + 2x + 2x^2$ - $\sin(x^2) \approx x^2$ - $\cos x \approx 1 - \frac{x^2}{2}$ - $e^{-x^3} \approx 1 - x^3$ - $\ln(2+x) - \ln 2 = \ln\left(\frac{2+x}{2}\right) = \ln\left(1 + \frac{x}{2}\right) \approx \frac{x}{2}$ - $e^{-x} \approx 1 - x$ - $x^2 e^{-x} \approx x^2 (1 - x) = x^2 - x^3$ - $\sin(x^2 e^{-x}) \approx \sin(x^2 - x^3) \approx x^2 - x^3$ - $\arctan\left(x + \frac{1}{x}\right)$ is not defined at $x=0$ because $\frac{1}{x}$ diverges. 4. **Check the denominator's behavior:** As $x \to 0$, $x + \frac{1}{x} \to \pm \infty$ depending on the direction, so $\arctan(x + \frac{1}{x}) \to \pm \frac{\pi}{2}$. 5. **Evaluate numerator near 0:** $$e^{2x} \sin(x^2) \approx (1 + 2x + 2x^2) x^2 = x^2 + 2x^3 + 2x^4$$ $$2 \cos x - 2 e^{-x^3} \approx 2 \left(1 - \frac{x^2}{2}\right) - 2 (1 - x^3) = 2 - x^2 - 2 + 2x^3 = -x^2 + 2x^3$$ Sum numerator: $$x^2 + 2x^3 + 2x^4 - x^2 + 2x^3 = 4x^3 + 2x^4$$ 6. **Evaluate denominator near 0:** $$(\ln(2+x) - \ln 2) \sin(x^2 e^{-x}) \arctan(x + \frac{1}{x}) \approx \frac{x}{2} (x^2 - x^3) \frac{\pi}{2} = \frac{\pi}{4} (x^3 - x^4)$$ 7. **Form the limit expression:** $$\lim_{x \to 0} \frac{4x^3 + 2x^4}{\frac{\pi}{4} (x^3 - x^4)} = \lim_{x \to 0} \frac{4x^3 + 2x^4}{\frac{\pi}{4} x^3 (1 - x)}$$ 8. **Cancel common factors:** $$= \lim_{x \to 0} \frac{\cancel{x^3} (4 + 2x)}{\frac{\pi}{4} \cancel{x^3} (1 - x)} = \lim_{x \to 0} \frac{4 + 2x}{\frac{\pi}{4} (1 - x)}$$ 9. **Simplify denominator fraction:** $$= \lim_{x \to 0} \frac{4 + 2x}{\frac{\pi}{4} (1 - x)} = \lim_{x \to 0} \frac{4 + 2x}{\frac{\pi}{4} - \frac{\pi}{4} x}$$ 10. **Evaluate limit by direct substitution:** $$= \frac{4 + 0}{\frac{\pi}{4} - 0} = \frac{4}{\frac{\pi}{4}} = 4 \times \frac{4}{\pi} = \frac{16}{\pi}$$ **Final answer:** $$\boxed{\frac{16}{\pi}}$$