1. **State the problem:** Find the limit $$\lim_{x \to 2} \frac{x-2}{x^2+4}$$. 2. **Recall the limit rule:** If direct substitution does not lead to an indeterminate form, substitute the value directly. 3. **Substitute $x=2$ into the expression:** $$\frac{2-2}{2^2+4} = \frac{0}{4+4} = \frac{0}{8} = 0$$ 4. **Conclusion:** The limit is $$0$$. This is because the numerator approaches 0 and the denominator approaches 8, so the fraction approaches 0.