1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{x(\ln(2+x-x^2)-\ln(2+x))\sin\bigl(x\sin(37+x^2)\bigr)}{(2x - x^2)\sin(x^2) - e^{2x^3} + 1}$$ 2. **Recall important formulas and rules:** - For small $x$, $\ln(1+u) \approx u$ when $u \to 0$. - $\sin y \approx y$ when $y \to 0$. - $e^z \approx 1 + z$ when $z \to 0$. - Use series expansions and simplifications to find the limit. 3. **Simplify the numerator:** $$\ln(2+x-x^2) - \ln(2+x) = \ln\left(\frac{2+x-x^2}{2+x}\right) = \ln\left(1 - \frac{x^2}{2+x}\right)$$ For small $x$, $2+x \approx 2$, so $$\ln\left(1 - \frac{x^2}{2+x}\right) \approx \ln(1 - \frac{x^2}{2}) \approx -\frac{x^2}{2}$$ Thus numerator becomes $$x \cdot \left(-\frac{x^2}{2}\right) \cdot \sin\bigl(x \sin(37 + x^2)\bigr) = -\frac{x^3}{2} \sin\bigl(x \sin(37 + x^2)\bigr)$$ 4. **Simplify $\sin\bigl(x \sin(37 + x^2)\bigr)$:** Since $x \to 0$, $x^2 \to 0$, so $$\sin(37 + x^2) \approx \sin(37) + x^2 \cos(37)$$ Therefore, $$x \sin(37 + x^2) \approx x \sin(37)$$ and $$\sin\bigl(x \sin(37 + x^2)\bigr) \approx \sin(x \sin(37)) \approx x \sin(37)$$ 5. **Substitute back into numerator:** $$-\frac{x^3}{2} \cdot x \sin(37) = -\frac{x^4}{2} \sin(37)$$ 6. **Simplify the denominator:** $$(2x - x^2) \sin(x^2) - e^{2x^3} + 1$$ For small $x$, $$\sin(x^2) \approx x^2$$ So, $$(2x - x^2) \sin(x^2) \approx (2x - x^2) x^2 = 2x^3 - x^4$$ Also, $$e^{2x^3} \approx 1 + 2x^3$$ Therefore denominator is $$2x^3 - x^4 - (1 + 2x^3) + 1 = 2x^3 - x^4 - 1 - 2x^3 + 1 = -x^4$$ 7. **Form the limit expression:** $$\lim_{x \to 0} \frac{-\frac{x^4}{2} \sin(37)}{-x^4} = \lim_{x \to 0} \frac{-\frac{x^4}{2} \sin(37)}{-x^4}$$ 8. **Cancel common factors:** $$= \lim_{x \to 0} \frac{\cancel{-x^4} \sin(37)/2}{\cancel{-x^4}} = \frac{\sin(37)}{2}$$ 9. **Final answer:** $$\boxed{\frac{\sin(37)}{2}}$$