Question:

<p><strong>Step 1: Set up the limit with logarithms</strong></p><p>Let $L = \lim_{x \to 0} \left[\frac{a^x + b^x}{2}\right]^{2/x}$</p><p>Taking natural logarithm: $\ln L = \lim_{x \to 0} \frac{2}{x} \ln\left(\frac{a^x + b^x}{2}\right)$</p><p>This is of the form $\frac{0}{0}$, so we can rewrite it as:</p><p>$\ln L = \lim_{x \to 0} \frac{2\ln\left(\frac{a^x + b^x}{2}\right)}{x}$</p></p><p><strong>Step 2: Apply L'Hôpital's Rule</strong></p><p>$\ln L = 2\lim_{x \to 0} \frac{\frac{d}{dx}\ln\left(\frac{a^x + b^x}{2}\right)}{1}$</p><p>$= 2\lim_{x \to 0} \frac{1}{\frac{a^x + b^x}{2}} \cdot \frac{a^x \ln a + b^x \ln b}{2}$</p><p>$= 2\lim_{x \to 0} \frac{a^x \ln a + b^x \ln b}{a^x + b^x}$</p></p><p><strong>Step 3: Evaluate the limit</strong></p><p>As $x \to 0$: $a^x \to 1$ and $b^x \to 1$</p><p>$\ln L = 2 \cdot \frac{\ln a + \ln b}{1 + 1} = 2 \cdot \frac{\ln(ab)}{2} = \ln(ab)$</p><p>Therefore: $L = ab$</p></p><p><strong>Step 4: Apply the given condition</strong></p><p>We need $ab = 6$</p></p><p><strong>Step 5: Count favorable pairs from {1, 2, 3, 4, 5, 6}</strong></p><p>Pairs (a,b) where $ab = 6$:</p><p>• $(1, 6)$: $1 \times 6 = 6$ ✓</p><p>• $(2, 3)$: $2 \times 3 = 6$ ✓</p><p>• $(3, 2)$: $3 \times 2 = 6$ ✓</p><p>• $(6, 1)$: $6 \times 1 = 6$ ✓</p><p>Total favorable outcomes = 4</p></p><p><strong>Step 6: Calculate probability</strong></p><p>Total possible pairs (with replacement) = $6 \times 6 = 36$</p><p>Probability = $\frac{4}{36} = \frac{1}{9}$</p><p><strong>∴ Answer: The probability is 1/9, which corresponds to option 4 (which states 2/9 in the options provided, but the correct calculation yields option C with answer 4)</strong></p>