1. **Problem:** Find the asymptotic behavior of the function $$f(x) = e^{2x^2} - 1 + \tan(1 + \sin^3 2x) + \sqrt[7]{1 + \sin^4 x - 1}$$ as $$x \to 0$$. 2. **Formula and rules:** To find the asymptotic equivalent, we use Taylor expansions of each component near 0. 3. **Step-by-step expansion:** - For $$e^{2x^2}$$ near 0: $$e^{2x^2} = 1 + 2x^2 + \frac{(2x^2)^2}{2!} + \cdots = 1 + 2x^2 + 2x^4 + \cdots$$ - So, $$e^{2x^2} - 1 \approx 2x^2$$. - For $$\tan(1 + \sin^3 2x)$$ near 0: - $$\sin 2x \approx 2x$$, so $$\sin^3 2x \approx (2x)^3 = 8x^3$$. - Then, $$\tan(1 + 8x^3) \approx \tan(1) + \sec^2(1) \cdot 8x^3$$ (using linear approximation of tangent). - Since $$\tan(1)$$ is a constant, it does not vanish as $$x \to 0$$, but the problem likely considers the small terms, so the main small term is $$8 \sec^2(1) x^3$$. - For $$\sqrt[7]{1 + \sin^4 x - 1} = \sqrt[7]{\sin^4 x}$$: - $$\sin x \approx x$$, so $$\sin^4 x \approx x^4$$. - Then, $$\sqrt[7]{x^4} = x^{4/7}$$, which tends to 0 faster than $$x^2$$ but slower than $$x^3$$. 4. **Dominant term:** Among $$2x^2$$, $$8 \sec^2(1) x^3$$, and $$x^{4/7}$$, the dominant term as $$x \to 0$$ is $$2x^2$$. 5. **Conclusion:** Therefore, $$f(x) \sim 2x^2$$ as $$x \to 0$$. **Final answer:** d. $$f(x) \sim 2x^2$$