1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{6e^x - 2x^2 - 4}{\sin(2x^2)}.$$\n\n2. **Recall important formulas and rules:**\n- The Taylor expansion of $e^x$ near 0 is $$e^x = 1 + x + \frac{x^2}{2} + \cdots.$$\n- The sine function near 0 behaves as $$\sin y \approx y$$ for small $y$.\n- We can use these approximations to simplify the expression and find the limit.\n\n3. **Apply the expansion to the numerator:**\n$$6e^x - 2x^2 - 4 = 6\left(1 + x + \frac{x^2}{2} + \cdots\right) - 2x^2 - 4 = 6 + 6x + 3x^2 - 2x^2 - 4 + \cdots = (6 - 4) + 6x + (3x^2 - 2x^2) + \cdots = 2 + 6x + x^2 + \cdots.$$\n\n4. **Apply the approximation to the denominator:**\n$$\sin(2x^2) \approx 2x^2$$ for small $x$.\n\n5. **Rewrite the limit using these approximations:**\n$$\lim_{x \to 0} \frac{2 + 6x + x^2 + \cdots}{2x^2}.$$\n\n6. **Analyze the behavior as $x \to 0$:**\nThe numerator approaches 2, but the denominator approaches 0, so the limit tends to infinity or does not exist as is. We need to check if the numerator also approaches 0 to get a finite limit.\n\n7. **Check the numerator at $x=0$:**\n$$6e^0 - 2(0)^2 - 4 = 6(1) - 0 - 4 = 2,$$ not zero, so the limit diverges.\n\n8. **Re-examine the problem:** The numerator does not approach zero, but the denominator does, so the limit is infinite. However, the problem likely expects a finite limit, so let's check if the original expression can be simplified differently.\n\n9. **Rewrite numerator as:**\n$$6e^x - 2x^2 - 4 = (6e^x - 6) + (6 - 2x^2 - 4) = 6(e^x - 1) + (2 - 2x^2).$$\n\n10. **Use expansion for $e^x - 1$:**\n$$e^x - 1 = x + \frac{x^2}{2} + \cdots,$$ so\n$$6(e^x - 1) = 6x + 3x^2 + \cdots.$$\n\n11. **Combine terms:**\n$$6x + 3x^2 + 2 - 2x^2 = 2 + 6x + x^2 + \cdots,$$ same as before.\n\n12. **Since numerator approaches 2 and denominator approaches 0, the limit is infinite.**\n\n13. **Check if the problem expects a limit of the form $\frac{0}{0}$ to apply L'Hôpital's Rule:**\nAt $x=0$, numerator = 2, denominator = 0, so no $0/0$ form.\n\n14. **Conclusion:** The limit does not exist (tends to infinity).\n\n**Final answer:** The limit does not exist (diverges to infinity).