We are tasked with solving the limit:

$\lim_{x \to 0} \left( \frac{1}{x} \left( \sqrt{\frac{1}{x^2} + 1} - \frac{1}{x} \right) + x \ln\left(1 + a^{1/x}\right) \right),$ where $a > 0$ and $a \neq 1$.

Step 1: Analyze the First Term

The first term is: $\frac{1}{x} \left( \sqrt{\frac{1}{x^2} + 1} - \frac{1}{x} \right).$ Simplifying the expression inside the square root: $\sqrt{\frac{1}{x^2} + 1} = \frac{\sqrt{1 + x^2}}{|x|}.$ For $x \to 0$, we approximate: $\sqrt{1 + x^2} \approx 1 + \frac{x^2}{2}.$ Thus: $\frac{1}{x} \left( \sqrt{\frac{1}{x^2} + 1} - \frac{1}{x} \right) \approx \frac{1}{x} \left( \frac{1 + \frac{x^2}{2}}{x} - \frac{1}{x} \right).$ This simplifies to: $\frac{1}{x} \left( \frac{1}{x} + \frac{x}{2} - \frac{1}{x} \right) = \frac{x}{2}.$ As $x \to 0$, this term tends to 0.

Step 2: Analyze the Second Term

The second term is: $x \ln\left(1 + a^{1/x}\right).$ As $x \to 0$, we can expand this term using logarithmic approximation. Since $a^{1/x}$ grows rapidly for small $x$, we know: $\ln\left(1 + a^{1/x}\right) \sim \ln\left(a^{1/x}\right) = \frac{1}{x} \ln a.$ Thus: $x \ln\left(1 + a^{1/x}\right) \approx x \cdot \frac{1}{x} \ln a = \ln a.$

Step 3: Combine the Results

The first term tends to 0, and the second term tends to $\ln a$. Therefore, the overall limit becomes: $\lim_{x \to 0} \left( 0 + \ln a \right) = \ln a.$

For the given choices:

• $\ln a = 1$ only when $a = e$, but $a$ was not specified as $e$, and the options do not suggest this.

Thus, the correct answer seems to be (D) None of these.

Would you like any additional clarification or details?

Follow-up questions:

1. What is the general approach to evaluating limits involving logarithmic terms?
2. How do approximations work in solving limits involving square roots?
3. Can you explain the approximation of $\ln(1 + a^{1/x})$ for small $x$?
4. What are other common methods to simplify expressions with indeterminate forms?
5. How does the behavior of $a^{1/x}$ as $x \to 0$ affect the result?

Tip:

When evaluating limits, using Taylor series expansions or approximations around small values of $x$ can help simplify complicated expressions.