1. **Problem f:** Evaluate $$\lim_{x \to 0} \frac{\frac{1}{x-6} + \frac{1}{6}}{3x}$$. Step 1: Simplify the numerator. $$\frac{1}{x-6} + \frac{1}{6} = \frac{6 + (x-6)}{6(x-6)} = \frac{x}{6(x-6)}$$. Step 2: Place it back in the limit expression. $$\lim_{x \to 0} \frac{\frac{x}{6(x-6)}}{3x} = \lim_{x \to 0} \frac{x}{6(x-6)} \cdot \frac{1}{3x} = \lim_{x \to 0} \frac{1}{6(x-6) \cdot 3} = \lim_{x \to 0} \frac{1}{18(x-6)}$$. Step 3: Substitute $x=0$. $$\frac{1}{18(0-6)} = \frac{1}{18 \times (-6)} = \frac{1}{-108} = -\frac{1}{108}$$. **Answer for f:** $$-\frac{1}{108}$$. 2. **Problem g:** Evaluate $$\lim_{x \to 0} \frac{\sqrt{x+6} - \sqrt{6}}{x}$$. Step 1: Multiply numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{x+6} - \sqrt{6}}{x} \times \frac{\sqrt{x+6} + \sqrt{6}}{\sqrt{x+6} + \sqrt{6}} = \frac{(x+6)-6}{x(\sqrt{x+6}+\sqrt{6})} = \frac{x}{x(\sqrt{x+6}+\sqrt{6})}$$. Step 2: Simplify by cancelling $x$ (for $x \neq 0$): $$\frac{1}{\sqrt{x+6}+\sqrt{6}}$$. Step 3: Take the limit as $x \to 0$: $$\frac{1}{\sqrt{6}+\sqrt{6}} = \frac{1}{2\sqrt{6}}$$. **Answer for g:** $$\frac{1}{2\sqrt{6}}$$. 3. **Problem h:** Evaluate $$\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1}}{7x}$$. Step 1: Simplify numerator: $$\sqrt{1+x} - 1$$. Step 2: Multiply numerator and denominator by the conjugate: $$\frac{\sqrt{1+x} - 1}{7x} \times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1} = \frac{(1+x) - 1}{7x(\sqrt{1+x}+1)} = \frac{x}{7x(\sqrt{1+x}+1)}$$. Step 3: Simplify by cancelling $x$ (for $x \neq 0$): $$\frac{1}{7(\sqrt{1+x}+1)}$$. Step 4: Take the limit as $x \to 0$: $$\frac{1}{7(1+1)} = \frac{1}{14}$$. **Answer for h:** $$\frac{1}{14}$$.