If a and b are chosen randomly from the set consisting of number 1, 2, 3, 4, 5, 6 with replacement. Then the probability that `lim_(x to 0)[(a^(x)+b^(x))//2]^(2//x)=6` is

Options

A. `1//3`
B. `1//4`
C. `1//9`
D. `2//9`

Explanation

To solve the problem, we need to determine the probability that

\[ \lim_{x \to 0} \left( \frac{a^x + b^x}{2} \right)^{\frac{2}{x}} = 6 \]

where \(a\) and \(b\) are chosen randomly from the set \{1, 2, 3, 4, 5, 6\} with replacement.

### Step 1: Analyze the limit expression

We start with the limit:

\[ \lim_{x \to 0} \left( \frac{a^x + b^x}{2} \right)^{\frac{2}{x}} \]

As \(x\) approaches 0, both \(a^x\) and \(b^x\) approach 1. Thus, we can rewrite the limit in a more manageable form:

\[ \lim_{x \to 0} \left( \frac{1 + 1}{2} \right)^{\frac{2}{x}} = \lim_{x \to 0} 1^{\frac{2}{x}} = 1 \]

However, we need to explore the expression further to find the conditions under which the limit equals 6.

### Step 2: Rewrite the limit using logarithms

Let:

\[ y = \lim_{x \to 0} \left( \frac{a^x + b^x}{2} \right)^{\frac{2}{x}} \]

Taking the natural logarithm of both sides, we have:

\[ \ln y = \lim_{x \to 0} \frac{2}{x} \ln \left( \frac{a^x + b^x}{2} \right) \]

### Step 3: Apply L'Hôpital's Rule

To evaluate this limit, we rewrite it as:

\[ \ln y = \lim_{x \to 0} \frac{2 \ln \left( \frac{a^x + b^x}{2} \right)}{x} \]

This is of the form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule:

\[ \ln y = \lim_{x \to 0} \frac{2 \cdot \frac{d}{dx} \left( \ln \left( \frac{a^x + b^x}{2} \right) \right)}{1} \]

Calculating the derivative:

\[ \frac{d}{dx} \left( \ln \left( \frac{a^x + b^x}{2} \right) \right) = \frac{a^x \ln a + b^x \ln b}{a^x + b^x} \]

### Step 4: Evaluate the limit

Now substituting back into the limit:

\[ \ln y = 2 \cdot \lim_{x \to 0} \frac{a^x \ln a + b^x \ln b}{a^x + b^x} \]

As \(x\) approaches 0, \(a^x\) and \(b^x\) both approach 1, hence:

\[ \ln y = 2 \cdot \frac{\ln a + \ln b}{2} = \ln(ab) \]

Thus, we have:

\[ y = ab \]

### Step 5: Set the condition for the limit

We want:

\[ ab = 6 \]

### Step 6: Find favorable outcomes

Now we need to find the pairs \((a, b)\) such that \(ab = 6\). The pairs that satisfy this condition from the set \{1, 2, 3, 4, 5, 6\} are:

1. (1, 6)
2. (2, 3)
3. (3, 2)
4. (6, 1)

Thus, there are 4 favorable outcomes.

### Step 7: Calculate total outcomes

Since \(a\) and \(b\) are chosen with replacement from 6 options, the total number of outcomes is:

\[ 6 \times 6 = 36 \]

### Step 8: Calculate the probability

The probability \(P\) that \(ab = 6\) is given by:

\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{36} = \frac{1}{9} \]

### Final Answer

Thus, the probability that

\[ \lim_{x \to 0} \left( \frac{a^x + b^x}{2} \right)^{\frac{2}{x}} = 6 \]

is

\[ \frac{1}{9} \]