Step 1: Understanding the Question:
We are asked to evaluate a limit, A, which is of the indeterminate form \(1^\infty\). After finding the value of A, we need to calculate its natural logarithm (log\(_e\) A).
Step 2: Key Formula or Approach:
For limits of the form \( \lim_{x \to a} [f(x)]^{g(x)} \) which result in \(1^\infty\), we can use the following formula:
\[ \lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} [f(x)-1]g(x)} \]
We will also use the standard trigonometric limit: \( \lim_{u \to 0} \frac{\tan u}{u} = 1 \).
Step 3: Detailed Explanation:
The given limit is \( A = \lim_{x \to 0^+} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}} \).
Here, \(f(x) = 1 + \tan^2 \sqrt{x}\) and \(g(x) = \frac{1}{2x}\).
As \(x \to 0^+\), \( \sqrt{x} \to 0 \), so \( \tan^2 \sqrt{x} \to 0 \). Thus, \(f(x) \to 1\).
As \(x \to 0^+\), \( g(x) = \frac{1}{2x} \to \infty \).
So, the limit is of the form \(1^\infty\).
Using the formula, we can write A as:
\[ A = e^L \]
where \( L = \lim_{x \to 0^+} [ (1 + \tan^2 \sqrt{x}) - 1 ] \times \frac{1}{2x} \)
\[ L = \lim_{x \to 0^+} [ \tan^2 \sqrt{x} ] \times \frac{1}{2x} \]
\[ L = \lim_{x \to 0^+} \frac{\tan^2 \sqrt{x}}{2x} \]
To evaluate this limit, we can manipulate it to use the standard limit \( \lim_{u \to 0} \frac{\tan u}{u} = 1 \).
Let's rewrite the expression for L:
\[ L = \frac{1}{2} \lim_{x \to 0^+} \frac{\tan^2 \sqrt{x}}{x} \]
\[ L = \frac{1}{2} \lim_{x \to 0^+} \left( \frac{\tan \sqrt{x}}{\sqrt{x}} \right)^2 \]
Let \(u = \sqrt{x}\). As \(x \to 0^+\), \(u \to 0^+\). The limit becomes:
\[ L = \frac{1}{2} \left( \lim_{u \to 0^+} \frac{\tan u}{u} \right)^2 \]
Since \( \lim_{u \to 0^+} \frac{\tan u}{u} = 1 \):
\[ L = \frac{1}{2} (1)^2 = \frac{1}{2} \]
So, the value of the original limit is \( A = e^L = e^{1/2} \).
The question asks for log\(_e\) A:
\[ \log_e A = \log_e (e^{1/2}) \]
Using the property \( \log_b (b^p) = p \):
\[ \log_e A = \frac{1}{2} \]
Step 4: Final Answer:
The value of log\(_e\) A is \( \frac{1}{2} \), which corresponds to option (C).