Question:

Step 1: Evaluate the limit using the standard form for indeterminate expressions. The limit $\lim_{x \to 0} \left(\frac{a^x + b^x}{2}\right)^{2/x}$ is of the form $1^\infty$. We use the standard result: $$\lim_{x \to 0} \left(\frac{a^x + b^x}{2}\right)^{2/x} = e^{\lim_{x\to 0} \frac{2}{x}\left(\frac{a^x+b^x}{2}-1\right)}$$ Step 2: Apply Taylor series expansion for exponential functions. Using the Taylor expansion $a^x = 1 + x\ln a + O(x^2)$ and $b^x = 1 + x\ln b + O(x^2)$, we get: $$\frac{a^x + b^x}{2} = \frac{(1 + x\ln a) + (1 + x\ln b)}{2} + O(x^2) = 1 + x \cdot \frac{\ln a + \ln b}{2} + O(x^2)$$ Step 3: Simplify the exponent of the exponential. Now we evaluate: $$\frac{2}{x}\left(\frac{a^x+b^x}{2}-1\right) = \frac{2}{x} \cdot x \cdot \frac{\ln a + \ln b}{2} + O(x) = \ln a + \ln b = \ln(ab)$$ As $x \to 0$, this approaches $\ln(ab)$. Step 4: Evaluate the limit. Substituting back into the exponential: $$\lim_{x \to 0} \left(\frac{a^x + b^x}{2}\right)^{2/x} = e^{\ln(ab)} = ab$$ Step 5: Determine the condition for the limit to equal 6. We need $ab = 6$. Now we find all pairs $(a,b)$ where both $a$ and $b$ are outcomes from a fair die (values from 1 to 6) and their product equals 6: - $(1, 6)$: $1 \times 6 = 6$ ✓ - $(6, 1)$: $6 \times 1 = 6$ ✓ - $(2, 3)$: $2 \times 3 = 6$ ✓ - $(3, 2)$: $3 \times 2 = 6$ ✓ There are **4 favorable outcomes**. Step 6: Calculate the probability. The total number of possible outcomes when throwing a pair of fair dice is: $$\text{Total outcomes} = 6 \times 6 = 36$$ Therefore, the probability is: $$P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{36} = \frac{1}{9}$$ **Final Answer:** The probability that $\lim_{x \to 0} \left(\frac{a^x + b^x}{2}\right)^{2/x} = 6$ equals $\boxed{\frac{1}{9}}$, which corresponds to **Option 4**.