1. Problem: Evaluate the limit $$\lim_{x \to 0} \left(1 + \frac{x}{a}\right)^{\frac{b}{x}}$$ where $a,b>0$. Step 1: Rewrite the expression inside the limit as an exponential \(e^{\ln}$. \[\lim_{x \to 0} \left(1 + \frac{x}{a}\right)^{\frac{b}{x}} = \lim_{x \to 0} e^{\frac{b}{x} \cdot \ln \left(1 + \frac{x}{a} \right)}\] Step 2: Consider the exponent $$\lim_{x \to 0} \frac{b}{x} \ln \left(1 + \frac{x}{a}\right)$$. Use the approximation \(\ln(1 + t) \approx t\) when \(t \to 0\). Here, \(t = \frac{x}{a}\). \[= \lim_{x \to 0} \frac{b}{x} \cdot \frac{x}{a} = \lim_{x \to 0} \frac{b}{a} = \frac{b}{a}\] Step 3: Therefore the limit is \[= e^{\frac{b}{a}}\] Answer: Choice (B) $e^{\frac{b}{a}}$. 2. Problem: Find $a$ such that $$f(x) = \frac{\log x}{x^2 - 1}$$ is continuous at $x=1$ when $f(1) = a$. Step 1: Continuity at $x=1$ means \[\lim_{x \to 1} f(x) = f(1) = a\] Step 2: Evaluate limit by noting numerator and denominator approach 0, use L'Hopital's Rule. \[\lim_{x \to 1} \frac{\log x}{x^2 - 1} = \lim_{x \to 1} \frac{1/x}{2x} = \lim_{x \to 1} \frac{1}{2x^2} = \frac{1}{2}\] Step 3: Set $a = \frac{1}{2}$ for continuity. Answer: Choice (B) $\frac{1}{2}$. 3. Problem: Probability that at least one of two lightbulbs chosen at random (without replacement) from 10 bulbs with 3 defective is defective. Step 1: Total ways to select 2 bulbs: $\binom{10}{2} = 45$. Step 2: Select 2 bulbs with none defective means choosing 2 from 7 good bulbs: $\binom{7}{2} = 21$. Step 3: Probability none defective: $\frac{21}{45} = \frac{7}{15}$. Step 4: Probability at least one defective: $1 - \frac{7}{15} = \frac{8}{15}$. Answer: Choice (C) $\frac{8}{15}$.