To solve the limit

$\lim_{x \to 0} \frac{x + \ln\left(\sqrt{x^2 + 1} - x\right)}{x^3}$

we can break it down and simplify it.

Step 1: Simplify the Logarithmic Term

First, consider the term inside the logarithm: $\sqrt{x^2 + 1} - x.$

Using the identity $\sqrt{a} - b = \frac{a - b^2}{\sqrt{a} + b}$ to simplify: $\sqrt{x^2 + 1} - x = \frac{(x^2 + 1) - x^2}{\sqrt{x^2 + 1} + x} = \frac{1}{\sqrt{x^2 + 1} + x}.$

Thus, $\ln\left(\sqrt{x^2 + 1} - x\right) = \ln\left(\frac{1}{\sqrt{x^2 + 1} + x}\right) = -\ln(\sqrt{x^2 + 1} + x).$

Step 2: Expand $\ln(\sqrt{x^2 + 1} + x)$ Around $x = 0$

Using the Taylor series for $\sqrt{x^2 + 1} \approx 1 + \frac{x^2}{2}$ as $x \to 0$: $\sqrt{x^2 + 1} + x \approx 1 + \frac{x^2}{2} + x.$

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

Using the Taylor expansion for $\ln(1 + u) \approx u$ when $u \to 0$: $\ln\left(1 + x + \frac{x^2}{2}\right) \approx x + \frac{x^2}{2}.$

Thus, $\ln\left(\sqrt{x^2 + 1} - x\right) \approx -(x + \frac{x^2}{2}).$

Step 3: Substitute and Simplify the Limit Expression

Now we rewrite the original limit expression using this approximation: $\lim_{x \to 0} \frac{x - x - \frac{x^2}{2}}{x^3} = \lim_{x \to 0} \frac{-\frac{x^2}{2}}{x^3}.$

This simplifies to: $\lim_{x \to 0} -\frac{x^2}{2x^3} = \lim_{x \to 0} -\frac{1}{2x} = 0.$

Therefore, the answer is:

$\boxed{0}.$

Would you like more details or have any questions?

Related Questions:

1. How can we apply L'Hôpital's Rule to solve similar indeterminate forms?
2. What are some alternative methods for simplifying limits with logarithmic terms?
3. How does the Taylor series expansion help in evaluating limits?
4. Can this approach be applied to higher-order terms in the Taylor series?
5. What other indeterminate forms commonly appear in limit problems?

Tip:

When dealing with limits that involve logarithms or square roots, simplifying by rationalizing or using series expansion often makes complex expressions more manageable.