1. **Problem statement:** Determine the value of the limit $$\lim_{x \to \pi} \frac{\tan^2 \left( \ln \left[ 1 + \sin (x - \pi) \right] \right)}{\sqrt[3]{1 + 3 \left(e^{x - \pi} - 1\right)^2}} - 1$$ 2. **Recall important formulas and rules:** - For small $t$, $\sin t \approx t$ and $e^t \approx 1 + t$. - The natural logarithm near 1: $\ln(1 + t) \approx t$ for small $t$. - The tangent function near 0: $\tan t \approx t$. - Cube root expansion: $\sqrt[3]{1 + u} \approx 1 + \frac{u}{3}$ for small $u$. 3. **Rewrite the limit using substitution:** Let $h = x - \pi$, so as $x \to \pi$, $h \to 0$. 4. **Evaluate inside the logarithm:** $$1 + \sin(h) \approx 1 + h$$ 5. **Evaluate the logarithm:** $$\ln(1 + \sin(h)) \approx \ln(1 + h) \approx h$$ 6. **Evaluate the tangent squared:** $$\tan^2(\ln(1 + \sin(h))) \approx \tan^2(h) \approx h^2$$ 7. **Evaluate the denominator inside the cube root:** $$e^h - 1 \approx h$$ $$3 (e^h - 1)^2 \approx 3 h^2$$ 8. **Evaluate the cube root:** $$\sqrt[3]{1 + 3 h^2} \approx 1 + \frac{3 h^2}{3} = 1 + h^2$$ 9. **Form the fraction inside the limit:** $$\frac{\tan^2(\ln(1 + \sin(h)))}{\sqrt[3]{1 + 3 (e^h - 1)^2}} \approx \frac{h^2}{1 + h^2}$$ 10. **Simplify the fraction:** $$\frac{h^2}{1 + h^2} = \frac{\cancel{h^2}}{1 + \cancel{h^2}}$$ 11. **Rewrite the original limit expression:** $$\lim_{h \to 0} \left( \frac{h^2}{1 + h^2} - 1 \right) = \lim_{h \to 0} \left( \frac{h^2 - (1 + h^2)}{1 + h^2} \right) = \lim_{h \to 0} \frac{h^2 - 1 - h^2}{1 + h^2} = \lim_{h \to 0} \frac{-1}{1 + h^2}$$ 12. **Evaluate the limit:** $$\lim_{h \to 0} \frac{-1}{1 + h^2} = \frac{-1}{1 + 0} = -1$$ **Final answer:** $$\boxed{-1}$$