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

First, let's rewrite the expression inside the natural logarithm:

$\ln\left(x + \sqrt{1 + x^2}\right)$

When $x \to 0$, we can use a Taylor expansion or approximation to simplify the expression inside the logarithm.

1. Simplifying the square root term:

$\sqrt{1 + x^2}$

For small $x$, we can use the approximation:

$\sqrt{1 + x^2} \approx 1 + \frac{x^2}{2}$

2. Substituting this approximation back into the expression:

$x + \sqrt{1 + x^2} \approx x + 1 + \frac{x^2}{2}$

Since $x$ is small, $x^2$ is even smaller, so the dominant terms are:

$x + 1$

3. Taking the natural logarithm:

$\ln\left(x + \sqrt{1 + x^2}\right) \approx \ln(1 + x)$

For $x$ close to 0, we can use the Taylor expansion of $\ln(1 + x)$:

$\ln(1 + x) \approx x$

Thus,

$\ln\left(x + \sqrt{1 + x^2}\right) \approx x$

So, as $x \to 0$,

$\ln\left(x + \sqrt{1 + x^2}\right) \sim x$

This completes the solution to the problem.

Would you like any further details or have any questions? Here are some related questions you might consider:

1. What is the Taylor series expansion for the natural logarithm function $\ln(1 + x)$?
2. How do you approximate expressions using Taylor series?
3. What are other methods for approximating limits of functions as $x \to 0$?
4. How does the approximation $\sqrt{1 + x^2} \approx 1 + \frac{x^2}{2}$ derive?
5. How can we apply these techniques to more complex functions?

Tip: When approximating functions near a specific point, Taylor series expansion is a powerful tool to simplify and understand the behavior of the function.