1. **State the problem:** Evaluate the limit $$\lim_{x \to 0} \frac{3x}{4 - 4\sec^2(5x)}.$$\n\n2. **Recall the formula and important rules:** The secant function is $$\sec(\theta) = \frac{1}{\cos(\theta)}$$ and near zero, $$\cos(\theta) \approx 1 - \frac{\theta^2}{2}$$ so $$\sec^2(\theta) = \frac{1}{\cos^2(\theta)}.$$\n\n3. **Rewrite the expression:**\n$$\frac{3x}{4 - 4\sec^2(5x)} = \frac{3x}{4(1 - \sec^2(5x))}.$$\n\n4. **Use the identity:** $$1 - \sec^2(\alpha) = -\tan^2(\alpha).$$\nSo denominator becomes $$4(1 - \sec^2(5x)) = 4(-\tan^2(5x)) = -4\tan^2(5x).$$\n\n5. **Rewrite the limit:**\n$$\lim_{x \to 0} \frac{3x}{-4\tan^2(5x)} = \lim_{x \to 0} \frac{3x}{-4(\tan(5x))^2}.$$\n\n6. **Approximate $$\tan(5x)$$ near zero:**\nSince $$\tan(\theta) \approx \theta$$ for small $$\theta$$,\n$$\tan(5x) \approx 5x.$$\n\n7. **Substitute approximation:**\n$$\lim_{x \to 0} \frac{3x}{-4(5x)^2} = \lim_{x \to 0} \frac{3x}{-4 \cdot 25 x^2} = \lim_{x \to 0} \frac{3x}{-100 x^2} = \lim_{x \to 0} \frac{3}{-100 x}.$$\n\n8. **Evaluate the limit:**\nAs $$x \to 0$$, $$\frac{3}{-100 x}$$ tends to $$\pm \infty$$ depending on the direction, so the limit does not exist as a finite number.\n\n**Final answer:** The limit does not exist (it diverges to infinity).