1. Problem: Compute $$\frac{f(x+h)-f(x)}{h}$$ at $$h=0$$ (finding the derivative as a limit) for the function $$f(x) = 3x^2 + 5$$. Step 1: Find $$f(x+h) = 3(x+h)^2 + 5 = 3(x^2 + 2xh + h^2) + 5 = 3x^2 + 6xh + 3h^2 + 5$$. Step 2: Compute $$\frac{f(x+h)-f(x)}{h} = \frac{3x^2 + 6xh + 3h^2 + 5 - (3x^2 + 5)}{h} = \frac{6xh + 3h^2}{h} = 6x + 3h$$. Step 3: Take the limit as $$h \to 0$$: $$\lim_{h \to 0} (6x + 3h) = 6x$$. 2. Problem: For $$f(x) = \frac{1}{x} + 1$$. Step 1: $$f(x+h) = \frac{1}{x+h} + 1$$. Step 2: $$\frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{x+h} + 1 - \left(\frac{1}{x} + 1\right)}{h} = \frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \frac{\frac{x - (x+h)}{x(x+h)}}{h} = \frac{\frac{-h}{x(x+h)}}{h} = -\frac{1}{x(x+h)}$$. Step 3: Limit as $$h \to 0$$: $$\lim_{h \to 0} -\frac{1}{x(x+h)} = -\frac{1}{x^2}$$. 3. Problem: For $$f(x) = \sqrt{x}$$. Step 1: $$f(x+h) = \sqrt{x+h}$$. Step 2: $$\frac{f(x+h)-f(x)}{h} = \frac{\sqrt{x+h} - \sqrt{x}}{h}$$. Rationalize numerator: $$\frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} = \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} = \frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \frac{1}{\sqrt{x+h} + \sqrt{x}}$$. Step 3: Limit as $$h \to 0$$: $$\lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$$. 4. Problem: For $$f(x) = x^3$$. Step 1: $$f(x+h) = (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$$. Step 2: $$\frac{f(x+h)-f(x)}{h} = \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} = \frac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2$$. Step 3: Limit as $$h \to 0$$: $$\lim_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2$$. 5. Problem: For $$f(x) = 5x^2 - 2x + 4$$. Step 1: $$f(x+h) = 5(x+h)^2 - 2(x+h) + 4 = 5(x^2 + 2xh + h^2) - 2x - 2h + 4 = 5x^2 + 10xh + 5h^2 - 2x - 2h + 4$$. Step 2: $$\frac{f(x+h)-f(x)}{h} = \frac{5x^2 + 10xh + 5h^2 - 2x - 2h + 4 - (5x^2 - 2x + 4)}{h} = \frac{10xh + 5h^2 - 2h}{h} = 10x + 5h - 2$$. Step 3: Limit as $$h \to 0$$: $$\lim_{h \to 0} (10x + 5h - 2) = 10x - 2$$.