If `a,b` are chosen from `{1,2,3,4,5,6,7}` randomly with replacement. The probability that `lim_(x to0) ((a^x+b^x)/2)^(2//x)=7 ` is

Options

A. `1//7`
B. `2//49`
C. `4//7`
D. `4//49`

Answer

To solve the problem, we need to find the probability that the limit

\[ \lim_{x \to 0} \left( \frac{a^x + b^x}{2} \right)^{\frac{2}{x}} = 7 \]

given that \(a\) and \(b\) are chosen randomly from the set \(\{1, 2, 3, 4, 5, 6, 7\}\) with replacement.

### Step-by-Step Solution:

1. **Understanding the Limit**:
We start with the expression inside the limit:
\[ \frac{a^x + b^x}{2} \]
As \(x\) approaches 0, both \(a^x\) and \(b^x\) approach 1 (since any number to the power of 0 is 1). Thus:
\[ \frac{a^x + b^x}{2} \to \frac{1 + 1}{2} = 1 \]

2. **Transforming the Limit**:
The limit can be rewritten as:
\[ \lim_{x \to 0} \left( \frac{a^x + b^x}{2} \right)^{\frac{2}{x}} = e^{\lim_{x \to 0} \frac{2}{x} \left( \frac{a^x + b^x}{2} - 1 \right)} \]
We need to evaluate the limit:
\[ \lim_{x \to 0} \frac{2}{x} \left( \frac{a^x + b^x}{2} - 1 \right) \]

3. **Using L'Hôpital's Rule**:
Since we have a \(0/0\) form, we can apply L'Hôpital's Rule. We differentiate the numerator and denominator:
- The derivative of the numerator \(2(a^x + b^x - 2)\) is:
\[ 2\left(a^x \ln a + b^x \ln b\right) \]
- The derivative of the denominator \(x\) is \(1\).

Thus, we rewrite the limit:
\[ \lim_{x \to 0} 2\left(a^x \ln a + b^x \ln b\right) \]

4. **Evaluating the Limit**:
As \(x \to 0\):
\[ a^x \to 1 \quad \text{and} \quad b^x \to 1 \]
Therefore, we have:
\[ \lim_{x \to 0} 2\left(1 \cdot \ln a + 1 \cdot \ln b\right) = 2(\ln a + \ln b) = \ln(ab^2) \]

5. **Setting the Limit Equal to 7**:
We want:
\[ e^{\ln(ab^2)} = 7 \implies ab^2 = 7 \]

6. **Finding Suitable Pairs (a, b)**:
We need to find pairs \((a, b)\) such that \(ab^2 = 7\). The possible values for \(a\) and \(b\) from the set \(\{1, 2, 3, 4, 5, 6, 7\}\) are:
- If \(a = 1\), then \(b^2 = 7\) (not possible).
- If \(a = 7\), then \(b^2 = 1\) (i.e., \(b = 1\)).
- If \(a = 7\) and \(b = 1\), we also have the pair \(a = 1\) and \(b = 7\).

The valid pairs are:
- \((1, 7)\)
- \((7, 1)\)

7. **Calculating the Probability**:
The total number of outcomes when choosing \(a\) and \(b\) is \(7 \times 7 = 49\). The favorable outcomes are \(2\) (the pairs found above).

Therefore, the probability is:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{49} \]

### Final Answer:
The probability that \(\lim_{x \to 0} \left( \frac{a^x + b^x}{2} \right)^{\frac{2}{x}} = 7\) is

\[ \frac{2}{49} \]