To evaluate the limit

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

we’ll analyze the expression in the numerator and check if simplifications can be applied.

Step 1: Simplifying $\sqrt{x^2 + 1} - x$

Rewrite $\sqrt{x^2 + 1} - x$ using a series expansion or rationalization. A Taylor expansion around $x = 0$ is a convenient approach here.

Using the Taylor expansion of $\sqrt{x^2 + 1}$ around $x = 0$: $\sqrt{x^2 + 1} = 1 + \frac{x^2}{2} + O(x^4).$ Thus, $\sqrt{x^2 + 1} - x \approx 1 + \frac{x^2}{2} - x.$

Step 2: Expanding $\ln(\sqrt{x^2 + 1} - x)$

Now, substitute this approximation into the logarithm term: $\ln(\sqrt{x^2 + 1} - x) = \ln\left(1 + \frac{x^2}{2} - x\right).$ Since $x$ is close to zero, we can expand this logarithm using $\ln(1 + u) \approx u$ when $u$ is small: $\ln\left(1 + \frac{x^2}{2} - x\right) \approx \frac{x^2}{2} - x.$

Step 3: Substituting back into the limit

Now, substitute back into the original limit: $\lim_{{x \to 0}} \frac{x + \left(\frac{x^2}{2} - x\right)}{x^3} = \lim_{{x \to 0}} \frac{x + \frac{x^2}{2} - x}{x^3}.$ The terms $x$ and $-x$ cancel out, leaving: $= \lim_{{x \to 0}} \frac{\frac{x^2}{2}}{x^3} = \lim_{{x \to 0}} \frac{x^2}{2x^3} = \lim_{{x \to 0}} \frac{1}{2x}.$

Step 4: Evaluating the limit

As $x \to 0$, $\frac{1}{2x}$ approaches infinity. Thus, the limit diverges to infinity.

$\lim_{{x \to 0}} \frac{x + \ln(\sqrt{x^2 + 1} - x)}{x^3} = \infty.$

The result is that the limit diverges to infinity.

Would you like a deeper breakdown or more questions on limits and Taylor expansions?

1. How does the Taylor expansion help in simplifying limits?
2. What is the purpose of logarithmic expansion in limit calculations?
3. How can rationalization be used to simplify limits involving square roots?
4. When do we apply L'Hôpital's rule, and how does it work here?
5. What are alternative methods to evaluate limits that go to infinity?

Tip: When faced with limits involving square roots and small $x$, try Taylor expansion or rationalization for efficient simplification.