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To evaluate the limit

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

we can proceed as follows:

### Step 1: Simplify the Expression Inside the Limit
Observe that \(a^x\) and \(b^x\) approach 1 as \(x \to 0\), since any non-zero number raised to the power of 0 is 1. Therefore, we have an indeterminate form of type \(1^{\infty}\), which suggests we should use logarithmic techniques to evaluate this limit.

### Step 2: Take the Natural Logarithm
Let \(L = \lim_{x \to 0} \left( \frac{a^x + b^x}{2} \right)^{\frac{1}{x}}\). Then we can rewrite this as

\[
\ln L = \lim_{x \to 0} \frac{1}{x} \ln \left( \frac{a^x + b^x}{2} \right).
\]

### Step 3: Apply L'Hôpital's Rule
The expression \(\frac{1}{x} \ln \left( \frac{a^x + b^x}{2} \right)\) is still indeterminate in the form \(\frac{0}{0}\) as \(x \to 0\), so we can apply L'Hôpital's Rule. Differentiating the numerator and denominator with respect to \(x\):

1. **Differentiate the Numerator:**  
   Using the chain rule, the derivative of \(\ln \left( \frac{a^x + b^x}{2} \right)\) is:
   \[
   \frac{1}{\frac{a^x + b^x}{2}} \cdot \frac{d}{dx} \left( \frac{a^x + b^x}{2} \right).
   \]
   Simplifying, this becomes:
   \[
   \frac{2 \cdot \left( a^x \ln a + b^x \ln b \right)}{2(a^x + b^x)} = \frac{a^x \ln a + b^x \ln b}{a^x + b^x}.
   \]

2. **Differentiate the Denominator:**  
   The derivative of \(x\) with respect to \(x\) is simply 1.

Thus, our limit becomes:

\[
\ln L = \lim_{x \to 0} \frac{a^x \ln a + b^x \ln b}{a^x + b^x}.
\]

### Step 4: Evaluate the Limit
As \(x \to 0\), \(a^x\) and \(b^x\) both approach 1. Substituting these limits, we get:

\[
\ln L = \frac{\ln a + \ln b}{2}.
\]

### Step 5: Exponentiate to Find \(L\)
Exponentiating both sides, we get:

\[
L = e^{\frac{\ln a + \ln b}{2}} = e^{\ln \sqrt{ab}} = \sqrt{ab}.
\]

### Conclusion
The evaluated limit is:

\[
\lim_{x \to 0} \left( \frac{a^x + b^x}{2} \right)^{\frac{1}{x}} = \sqrt{ab}.
\]

Would you like further details on any part of this solution?

### Related Questions
1. How do you evaluate limits that involve indeterminate forms like \(1^{\infty}\)?
2. What are some other techniques for limits involving logarithmic functions?
3. How can L'Hôpital's Rule be applied in more complex expressions?
4. Why do exponential functions simplify to 1 as their exponents approach 0?
5. How does the limit process change if we have different types of indeterminate forms?

### Tip
For limits involving indeterminate forms, especially exponential functions, logarithmic transformation can simplify the problem significantly.

Evaluate the following limit, lim x→0 ((a^x + b^x) / 2)^(1/x).

Learn how to evaluate the limit lim x→0 ((a^x + b^x) / 2)^(1/x) using logarithmic techniques and L'Hôpital's Rule. This solution addresses an indeterminate form and applies logarithmic transformations for simplification, making it suitable for undergraduate mathematics students.

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