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(D) Let $L = \mathop {\lim }\limits_{x 	o 0} {\left( {\frac{{{a^x} + {b^x}}}{2}} ight)^{\frac{2}{x}}}$.Since the limit is of the form $1^\infty$,we

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$\frac{1}{9}$

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(D) Let $L = \mathop {\lim }\limits_{x 	o 0} {\left( {\frac{{{a^x} + {b^x}}}{2}} ight)^{\frac{2}{x}}}$.Since the limit is of the form $1^\infty$,we use the formula $\mathop {\lim }\limits_{x 	o 0} f(x)^{g(x)} = e^{\mathop {\lim }\limits_{x 	o 0} g(x)(f(x)-1)}$.$L = e^{\mathop {\lim }\limits_{x 	o 0} \frac{2}{x} \left( \frac{a^x + b^x}{2} - 1 ight)} = e^{\mathop {\lim }\limits_{x 	o 0} \frac{a^x + b^x - 2}{x}}$.Using the standard limit $\mathop {\lim }\limits_{x 	o 0} \frac{k^x - 1}{x} = \ln k$,we have:$L = e^{\mathop {\lim }\limits_{x 	o 0} \left( \frac{a^x - 1}{x} + \frac{b^x - 1}{x} ight)} = e^{\ln a + \ln b} = e^{\ln(ab)} = ab$.We are given $ab = 6$.The possible pairs $(a, b)$ from the set $\{1, 2, 3, 4, 5, 6\}$ such that $ab = 6$ are $(1, 6), (6, 1), (2, 3), (3, 2)$.Total number of outcomes when picking two numbers with replacement is $6 	imes 6 = 36$.Number of favorable outcomes is $4$.Probability = $\frac{4}{36} = \frac{1}{9}$.

If $a$ and $b$ are chosen randomly from the set $\{1, 2, 3, 4, 5, 6\}$ with replacement,then the probability that $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x}}}{2}} \right)^{\frac{2}{x}}}=6$ is

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