1. We need to find the limit $$\lim_{x \to \infty} \left(1 + \frac{x^b}{a}\right)^{\frac{a}{x}}$$ where $a,b>0$. 2. Rewrite inside the power using exponential and logarithm: $$ = \lim_{x \to \infty} e^{\frac{a}{x} \ln\left(1 + \frac{x^b}{a}\right)} $$ 3. Analyze the exponent: $$ \frac{a}{x} \ln\left(1 + \frac{x^b}{a}\right) = a \cdot \frac{\ln\left(1 + \frac{x^b}{a}\right)}{x} $$ 4. For large $x$, $\frac{x^b}{a}$ dominates inside the log, so: $$ \ln\left(1 + \frac{x^b}{a}\right) \sim \ln\left(\frac{x^b}{a}\right) = b \ln x - \ln a $$ 5. Substitute into exponent: $$ a \cdot \frac{b \ln x - \ln a}{x} = a b \frac{\ln x}{x} - a \frac{\ln a}{x} $$ 6. As $x \to \infty$, $\frac{\ln x}{x} \to 0$ and $\frac{\ln a}{x} \to 0$, so the exponent tends to 0. 7. Therefore, the limit is: $$ e^0 = 1 $$ **Answer for 7:** (D) 1 --- 1. Given $f(x) = \frac{\log x}{x^2 - 1}$ for $x > 0, x \neq 1$, and define $f(1) = a$, find $a$ for continuity at $x=1$. 2. Continuity at $x=1$ requires: $$ \lim_{x \to 1} f(x) = f(1) = a $$ 3. Compute: $$ \lim_{x \to 1} \frac{\log x}{x^2 - 1} = \lim_{x \to 1} \frac{\log x}{(x-1)(x+1)} $$ 4. Direct substitution gives $\frac{0}{0}$, apply L'Hôpital's Rule. 5. Differentiate numerator and denominator with respect to $x$: Numerator derivative: $\frac{1}{x}$ Denominator derivative: $2x$ 6. Limit becomes: $$ \lim_{x \to 1} \frac{1/x}{2x} = \lim_{x \to 1} \frac{1}{2x^2} = \frac{1}{2} $$ 7. Thus, to be continuous at $x=1$, we need: $$ a = \frac{1}{2} $$ **Answer for 8:** (B) 1/2 --- 1. Box has 10 lightbulbs, 3 defective, 7 good. 2. Two bulbs chosen without replacement. Find probability at least 1 defective. 3. Use complement: probability (at least 1 defective) = 1 - probability (none defective). 4. Probability none defective (both good): $$ \frac{7}{10} \times \frac{6}{9} = \frac{42}{90} = \frac{7}{15} $$ 5. Therefore, $$ P(\text{at least 1 defective}) = 1 - \frac{7}{15} = \frac{8}{15} $$ **Answer for 9:** (C) 8/15