To solve the limit using L'Hôpital's rule, let’s analyze the given limit step by step:

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

Step 1: Simplify and analyze the structure

• The key term here is $x^2 \ln\left(1 + \frac{1}{x}\right)$, which involves a logarithm that can behave singularly as $x \to 0$.
• Split the limit as follows:

$L = \lim_{x \to 0} \left( x - x^2 \ln\left(1 + \frac{1}{x}\right) \right).$

Since the limit is indeterminate of the form $0 \cdot \infty$, it’s ideal to apply L'Hôpital's rule after rewriting the expression.

Step 2: Analyze $\ln\left(1 + \frac{1}{x}\right)$

For $x \to 0$, we note that $\frac{1}{x} \to \infty$, and therefore:

$\ln\left(1 + \frac{1}{x}\right) \sim \ln\left(\frac{1}{x}\right) \quad \text{as} \ x \to 0.$

Thus, we approximate:

$\ln\left(1 + \frac{1}{x}\right) \approx \ln\left(\frac{1}{x}\right) = -\ln(x).$

Step 3: Substitute the approximation

Replace $\ln\left(1 + \frac{1}{x}\right)$ with $-\ln(x)$:

$L = \lim_{x \to 0} \left( x - x^2 \cdot (-\ln(x)) \right).$

Simplify the expression:

$L = \lim_{x \to 0} \left( x + x^2 \ln(x) \right).$

Here:

1. $x \to 0$ is straightforward.
2. $x^2 \ln(x)$ involves an indeterminate form $0 \cdot -\infty$, so we isolate it for L'Hôpital’s rule.

Step 4: Focus on $x^2 \ln(x)$ using L'Hôpital's Rule

Consider the term:

$\lim_{x \to 0} x^2 \ln(x).$

To apply L'Hôpital's rule, rewrite it as:

$\lim_{x \to 0} x^2 \ln(x) = \lim_{x \to 0} \frac{\ln(x)}{\frac{1}{x^2}}.$

Now, this is an indeterminate form $\frac{-\infty}{\infty}$. We differentiate the numerator and denominator:

• Derivative of $\ln(x)$ is $\frac{1}{x}$,
• Derivative of $\frac{1}{x^2}$ is $-\frac{2}{x^3}$.

Thus:

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

Simplify:

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

Step 5: Combine the results

Since $x^2 \ln(x) \to 0$ as $x \to 0$, we have:

$L = \lim_{x \to 0} \left( x + 0 \right) = 0.$

Final Answer:

$\boxed{0}$

Do you need a more detailed explanation for any step, or additional examples on L'Hôpital's rule? Let me know!

Related Questions:

1. What are the conditions for applying L'Hôpital's rule?
2. How do you evaluate limits involving $x \ln(x)$ as $x \to 0$?
3. What is the behavior of $\ln(x)$ near $x = 0$?
4. How do you approximate $\ln(1 + x)$ for small $x$?
5. Can L'Hôpital's rule be applied multiple times?

Tip: For limits involving logarithmic and polynomial terms, logarithmic approximations and L'Hôpital's rule are powerful tools.