1. **State the problem:** We need to find the limit as $n$ approaches infinity of the expression $$\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + \cdots - 2n}{\sqrt{n^2 + 1} + \sqrt{4n^2 + 1}}.$$\n\n2. **Simplify the numerator:** The numerator is the sum of the first $2n$ terms of the sequence with alternating signs: $$1 - 2 + 3 - 4 + \cdots + (2n-1) - 2n.$$\nGroup the terms in pairs: $$(1 - 2) + (3 - 4) + \cdots + ((2n-1) - 2n).$$\nEach pair sums to $-1$, and there are $n$ such pairs, so the numerator equals $$n \times (-1) = -n.$$\n\n3. **Simplify the denominator:** The denominator is $$\sqrt{n^2 + 1} + \sqrt{4n^2 + 1}.$$\nAs $n \to \infty$, use the approximation for large $n$: $$\sqrt{a n^2 + b} \approx n\sqrt{a} + \frac{b}{2n\sqrt{a}}.$$\nApplying this to each term: $$\sqrt{n^2 + 1} \approx n + \frac{1}{2n},$$ $$\sqrt{4n^2 + 1} \approx 2n + \frac{1}{4n}.$$\nSumming gives $$n + \frac{1}{2n} + 2n + \frac{1}{4n} = 3n + \frac{3}{4n}.$$\n\n4. **Form the fraction:** The expression is approximately $$\frac{-n}{3n + \frac{3}{4n}}.$$\nDivide numerator and denominator by $n$ to simplify: $$\frac{-n/n}{(3n / n) + \frac{3}{4n} \times \frac{1}{n/n}} = \frac{-1}{3 + \frac{3}{4n^2}}.$$\n\n5. **Take the limit:** As $n \to \infty,$ the term $\frac{3}{4n^2} \to 0,$ so the limit becomes $$\lim_{n \to \infty} \frac{-1}{3 + 0} = -\frac{1}{3}.$$\n\n**Final answer:** $$\boxed{-\frac{1}{3}}.$$