1. **State the problem:** We need to find the limit $$\lim_{h \to 0} \frac{\frac{5}{(2+h)^2} - \frac{5}{2^2}}{h}$$ 2. **Recognize the form:** This is the definition of the derivative of the function $$f(x) = \frac{5}{x^2}$$ at the point $x=2$. 3. **Recall the derivative formula:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 4. **Apply the function:** $$f(2+h) = \frac{5}{(2+h)^2}, \quad f(2) = \frac{5}{2^2} = \frac{5}{4}$$ 5. **Rewrite the limit:** $$\lim_{h \to 0} \frac{\frac{5}{(2+h)^2} - \frac{5}{4}}{h}$$ 6. **Find a common denominator inside the numerator:** $$\frac{5}{(2+h)^2} - \frac{5}{4} = \frac{5 \cdot 4 - 5 \cdot (2+h)^2}{4 (2+h)^2} = \frac{20 - 5(2+h)^2}{4 (2+h)^2}$$ 7. **Substitute back into the limit:** $$\lim_{h \to 0} \frac{\frac{20 - 5(2+h)^2}{4 (2+h)^2}}{h} = \lim_{h \to 0} \frac{20 - 5(2+h)^2}{4 (2+h)^2 \cdot h}$$ 8. **Expand $(2+h)^2$:** $$(2+h)^2 = 4 + 4h + h^2$$ 9. **Substitute expansion:** $$20 - 5(4 + 4h + h^2) = 20 - 20 - 20h - 5h^2 = -20h - 5h^2$$ 10. **Rewrite the limit:** $$\lim_{h \to 0} \frac{-20h - 5h^2}{4 (4 + 4h + h^2) h}$$ 11. **Factor $h$ from numerator:** $$\lim_{h \to 0} \frac{h(-20 - 5h)}{4 (4 + 4h + h^2) h}$$ 12. **Cancel $h$ using \cancel:** $$\lim_{h \to 0} \frac{\cancel{h}(-20 - 5h)}{4 (4 + 4h + h^2) \cancel{h}}$$ 13. **Simplify:** $$\lim_{h \to 0} \frac{-20 - 5h}{4 (4 + 4h + h^2)}$$ 14. **Evaluate the limit by substituting $h=0$:** $$\frac{-20 - 0}{4 (4 + 0 + 0)} = \frac{-20}{16} = -\frac{5}{4}$$ **Final answer:** $$\boxed{-\frac{5}{4}}$$