The value of $$\lim_{n \to \infty} \frac{1+2-3+4+5-6+\ldots+(3n-2)+(3n-1)-3n}{\sqrt{2n^4+4n+3} - \sqrt{n^4+5n+4}}$$ is

$$S_n = 1 + 2 - 3 + 4 + 5 - 6 + \dots + (3n-2) + (3n-1) - 3n$$

Each triplet $$(3k-2) + (3k-1) - 3k$$ simplifies to $$3k - 3$$

$$S_n = \sum_{k=1}^n (3k - 3) = 3 \sum_{k=1}^n (k-1) = 3 \left( \frac{(n-1)n}{2} \right) = \frac{3n(n-1)}{2}$$

The denominator is $$D_n = \sqrt{2n^4 + 4n + 3} - \sqrt{n^4 + 5n + 4}$$.

As $$n \to \infty$$, we can factor out $$n^2$$ from each square root:
$$D_n = n^2 \sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}} - n^2 \sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}}$$
$$D_n \approx n^2 (\sqrt{2} - 1)$$

$$\lim_{n\to\infty} \frac{\frac{3n^2 - 3n}{2}}{n^2(\sqrt{2} - 1)}$$

$$\lim_{n\to\infty} \frac{\frac{3}{2} - \frac{3}{2n}}{\sqrt{2} - 1} = \frac{3}{2(\sqrt{2} - 1)}$$

$$\frac{3(\sqrt{2} + 1)}{2(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{3(\sqrt{2} + 1)}{2(2 - 1)} = \frac{3}{2}(\sqrt{2} + 1)$$