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To solve the given problem, we need to determine the limit of \(\ln(x + \sqrt{1 + x^2})\) as \(x \to 0\).

Let's begin by examining the expression inside the logarithm.

Given: 
\[
\ln\left(x + \sqrt{1 + x^2}\right)
\]

As \(x \to 0\), we can use the first-order Taylor expansion for small \(x\). For small \(x\), \(\sqrt{1 + x^2}\) can be approximated as follows:
\[
\sqrt{1 + x^2} \approx 1 + \frac{x^2}{2}
\]

So, the expression inside the logarithm becomes:
\[
x + \sqrt{1 + x^2} \approx x + 1 + \frac{x^2}{2}
\]

Now, let's consider the logarithm:
\[
\ln\left(x + 1 + \frac{x^2}{2}\right)
\]

For \(x\) very small, we can use the Taylor expansion of \(\ln(1 + y) \approx y\) when \(y \to 0\).

Here, \(y = x + \frac{x^2}{2}\) is also very small as \(x \to 0\).

So,
\[
\ln\left(1 + x + \frac{x^2}{2}\right) \approx x + \frac{x^2}{2}
\]

Thus, as \(x \to 0\),
\[
\ln\left(x + \sqrt{1 + x^2}\right) \sim x
\]

Therefore, the limit is:
\[
\lim_{x \to 0} \ln\left(x + \sqrt{1 + x^2}\right) = x
\]

If you have any questions or need further details, please let me know.

Here are 5 questions you might consider asking next:
1. Can you explain the Taylor expansion used in this solution?
2. How would the solution change if \(x\) approaches a different value, like infinity?
3. Can you provide more examples of logarithmic limits?
4. How do you approach solving limits involving square roots in general?
5. What are some common mistakes to avoid when dealing with limits and logarithms?

Tip: When dealing with limits, using Taylor series expansions can be very helpful for approximating functions near specific points.

Learn how to solve the limit of ln(x + sqrt(1 + x^2)) as x approaches 0 with a detailed step-by-step solution using Taylor series approximation and logarithmic limit principles. Suitable for advanced high school students and beyond.

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