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.