The given problem involves evaluating the following limit:

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

Solution:

Step 1: Simplify the denominator

The denominator $x - 1$ approaches $-1$ as $x \to 0^+$.

Step 2: Analyze the numerator

The numerator $x \left(\ln(x)\right)^2$ behaves as follows:

• As $x \to 0^+$, $\ln(x) \to -\infty$, so $\left(\ln(x)\right)^2 \to +\infty$.
• The $x$ factor approaches $0$.

Thus, the numerator is an indeterminate form of type $0 \cdot \infty$.

Step 3: Rewriting for evaluation

We can rewrite the numerator as: $x \left(\ln(x)\right)^2 = \frac{\left(\ln(x)\right)^2}{1/x}.$ Thus, the limit becomes: $\lim_{x \to 0^+} \frac{\left(\ln(x)\right)^2}{\frac{1}{x}(x-1)}.$

Step 4: Approximation for small $x$

When $x \to 0^+$, $\frac{1}{x}(x-1) \to -\frac{1}{x}$, so the limit simplifies to: $\lim_{x \to 0^+} \frac{\left(\ln(x)\right)^2}{-\frac{1}{x}} = -x \left(\ln(x)\right)^2.$

Now, as $x \to 0^+$:

• $-x$ approaches $0$,
• and $\left(\ln(x)\right)^2 \to \infty$.

Thus, the product remains $0 \cdot \infty$, requiring further analysis.

Step 5: Applying L'Hôpital's Rule

Express the limit as a fraction: $\lim_{x \to 0^+} \frac{\left(\ln(x)\right)^2}{1/x}.$ This is now an indeterminate form $\frac{\infty}{\infty}$. Applying L’Hôpital's Rule: $\frac{d}{dx} \left(\left(\ln(x)\right)^2\right) = 2\ln(x) \cdot \frac{1}{x}, \quad \frac{d}{dx} \left(\frac{1}{x}\right) = -\frac{1}{x^2}.$ Thus, the limit becomes: $\lim_{x \to 0^+} \frac{2\ln(x) \cdot \frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} \frac{-2\ln(x)}{1/x} = \lim_{x \to 0^+} -2x\ln(x).$

Step 6: Evaluate the remaining limit

Rewriting $x \ln(x)$ as $\frac{\ln(x)}{1/x}$, this approaches $0$ as $x \to 0^+$.

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

Would you like me to expand or clarify any steps?

Related Questions:

1. What is the intuition behind L’Hôpital's Rule, and when can it be applied?
2. How does the logarithmic function $\ln(x)$ behave as $x \to 0^+$?
3. Why is $0 \cdot \infty$ considered an indeterminate form?
4. Can we evaluate this limit numerically for small $x$ to confirm the result?
5. What are some common limits involving logarithmic functions?

Tip:

When evaluating indeterminate forms, always consider rewriting the expression and check if L’Hôpital's Rule applies before proceeding!