1. We are asked to find the limit $$\lim_{h \to 0} \frac{f(8+h) - f(8)}{h}$$ where $$f(x) = x^2 + 3$$. 2. This limit represents the definition of the derivative of $$f(x)$$ at $$x=8$$. 3. First, compute $$f(8+h)$$: $$f(8+h) = (8+h)^2 + 3 = 64 + 16h + h^2 + 3 = 67 + 16h + h^2$$. 4. Compute $$f(8)$$: $$f(8) = 8^2 + 3 = 64 + 3 = 67$$. 5. Substitute into the difference quotient: $$\frac{f(8+h) - f(8)}{h} = \frac{(67 + 16h + h^2) - 67}{h} = \frac{16h + h^2}{h}$$. 6. Simplify the fraction by canceling $$h$$: $$\frac{\cancel{h}(16 + h)}{\cancel{h}} = 16 + h$$. 7. Take the limit as $$h \to 0$$: $$\lim_{h \to 0} (16 + h) = 16$$. 8. Therefore, the limit is $$16$$, which is the derivative of $$f(x)$$ at $$x=8$$.