1. Consider the limit \( \lim_{x \to 0} \left(1 + \frac{x}{a}\right)^{\frac{b}{x}} \) where \(a,b > 0\).\n 2. Rewrite the expression using the exponential and logarithm: \n\[ \left(1 + \frac{x}{a}\right)^{\frac{b}{x}} = e^{\frac{b}{x} \ln \left(1 + \frac{x}{a}\right)} \]\n 3. Focus on the exponent: \n\[ \frac{b}{x} \ln \left(1 + \frac{x}{a}\right) \]\n 4. For small \(x\), use the approximation \( \ln(1 + y) \approx y \), so \( \ln \left(1 + \frac{x}{a}\right) \approx \frac{x}{a} \).\n 5. Substitute into the exponent: \n\[ \frac{b}{x} \cdot \frac{x}{a} = \frac{b}{a} \]\n 6. Therefore, the limit becomes: \n\[ \lim_{x \to 0} e^{\frac{b}{a}} = e^{\frac{b}{a}} \]\n 7. So the answer is \( e^{\frac{b}{a}} \), which corresponds to option (B).\n 8. Now consider the function \( f(x) = \frac{\log x}{x^2 - 1} \) for \( x > 0 \), \( x \neq 1 \), and define \( f(1) = a \).\n 9. To ensure continuity at \( x = 1 \), \( \lim_{x \to 1} f(x) = f(1) = a \).\n 10. Find \( \lim_{x\to1} \frac{\log x}{x^2 - 1} = \lim_{x\to1} \frac{\log x}{(x-1)(x+1)} \).\n 11. Since this is \( \frac{0}{0} \) form, apply L'Hôpital's Rule differentiating numerator and denominator w.r.t. \( x \):\n\[ \lim_{x \to 1} \frac{\frac{1}{x}}{2x} = \lim_{x\to1} \frac{1/x}{2x} = \frac{1/1}{2 \cdot 1} = \frac{1}{2} \]\n 12. Hence, for continuity, \( a = \frac{1}{2} \), option (B).\n 13. For the probability question: 3 out of 10 bulbs are defective, pick 2 without replacement. Find probability at least one is defective.\n 14. Calculate complement: probability none defective (both good). Number of good bulbs = 7.\n 15. Number of ways to pick 2 good bulbs: \( \binom{7}{2} = 21 \). Total ways to pick any 2 bulbs: \( \binom{10}{2} = 45 \).\n 16. Probability no defective bulbs is \( \frac{21}{45} = \frac{7}{15} \).\n 17. Thus probability at least one defective: \n\[ 1 - \frac{7}{15} = \frac{8}{15} \]\n 18. Answer is option (C).