1. We are asked to find the limit as $n$ approaches 0 of the expression $$\frac{1-\cos 2n}{n}$$. 2. Recall the trigonometric identity: $$1-\cos x = 2\sin^2 \frac{x}{2}$$. 3. Applying this identity with $x=2n$, we get: $$\frac{1-\cos 2n}{n} = \frac{2\sin^2 n}{n}$$. 4. Now the limit becomes: $$\lim_{n \to 0} \frac{2\sin^2 n}{n}$$. 5. We can rewrite $\sin^2 n$ as $(\sin n)^2$ and use the fact that $\lim_{n \to 0} \frac{\sin n}{n} = 1$. 6. Express the limit as: $$\lim_{n \to 0} 2 \sin n \cdot \frac{\sin n}{n}$$. 7. Since $\lim_{n \to 0} \sin n = 0$ and $\lim_{n \to 0} \frac{\sin n}{n} = 1$, the product limit is: $$2 \times 0 \times 1 = 0$$. 8. Therefore, the limit is: $$\boxed{0}$$.