1. **State the problem:** We want to find the limit $$\lim_{x \to 0} \frac{1 - \cos^2 x}{5x}.$$\n\n2. **Recall the identity:** Note that $$1 - \cos^2 x = \sin^2 x$$ by the Pythagorean identity. So the limit becomes $$\lim_{x \to 0} \frac{\sin^2 x}{5x}.$$\n\n3. **Rewrite the expression:** We can write $$\frac{\sin^2 x}{5x} = \frac{\sin x}{x} \cdot \frac{\sin x}{5}.$$\n\n4. **Use the standard limit:** We know that $$\lim_{x \to 0} \frac{\sin x}{x} = 1.$$\n\n5. **Evaluate the limit:** Therefore, $$\lim_{x \to 0} \frac{\sin^2 x}{5x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \frac{\sin x}{5} = 1 \cdot \frac{0}{5} = 0.$$\n\n**Final answer:** $$\boxed{0}.$$