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To evaluate the limit \(\lim_{x \to 0^+} \sqrt[5]{x} \ln(x)\), let's analyze the behavior of the function as \(x\) approaches \(0^+\) from the positive side.

### Step-by-Step Solution:

1. **Understanding the components**:
   - \(\sqrt[5]{x} = x^{1/5}\), which approaches \(0\) as \(x \to 0^+\).
   - \(\ln(x)\) approaches \(-\infty\) as \(x \to 0^+\).

2. **Combining the components**:
   - The expression \(\sqrt[5]{x} \ln(x) = x^{1/5} \ln(x)\) involves a product of \(x^{1/5}\) and \(\ln(x)\).
   - As \(x \to 0^+\), \(x^{1/5} \to 0\) and \(\ln(x) \to -\infty\). The limit results in the indeterminate form \(0 \cdot (-\infty)\).

3. **Transforming to a more manageable form**:
   - To resolve the indeterminate form, rewrite the expression as:
   \[
   \lim_{x \to 0^+} x^{1/5} \ln(x) = \lim_{x \to 0^+} \frac{\ln(x)}{x^{-1/5}}.
   \]
   - This transforms the expression into a \(\frac{-\infty}{\infty}\) indeterminate form.

4. **Applying L'Hopital's Rule**:
   - To apply L'Hopital's Rule, differentiate the numerator and denominator:
   \[
   \text{Numerator: } \frac{d}{dx}[\ln(x)] = \frac{1}{x},
   \]
   \[
   \text{Denominator: } \frac{d}{dx}[x^{-1/5}] = -\frac{1}{5}x^{-6/5}.
   \]
   - Applying L'Hopital's Rule gives:
   \[
   \lim_{x \to 0^+} \frac{\ln(x)}{x^{-1/5}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{5}x^{-6/5}} = \lim_{x \to 0^+} -\frac{5x^{6/5}}{x} = \lim_{x \to 0^+} -5x^{1/5}.
   \]

5. **Evaluating the simplified limit**:
   - Now, evaluate the limit:
   \[
   \lim_{x \to 0^+} -5x^{1/5} = -5 \cdot 0 = 0.
   \]

Therefore, the limit is:
\[
\lim_{x \to 0^+} \sqrt[5]{x} \ln(x) = 0.
\]

Would you like further details or have any questions about this solution?

### Related Questions:

1. How would the solution change if the limit was \(\lim_{x \to \infty} \sqrt[5]{x} \ln(x)\)?
2. Can we evaluate \(\lim_{x \to 0^+} x^a \ln(x)\) for any other values of \(a\)?
3. What are the conditions for applying L'Hopital's Rule?
4. How does the behavior of \(\ln(x)\) as \(x \to 0^+\) differ from its behavior as \(x \to \infty\)?
5. What other methods can be used to evaluate limits involving indeterminate forms besides L'Hopital's Rule?

### Tip:
When encountering limits involving indeterminate forms like \(0 \cdot (-\infty)\), consider transforming the expression into a fraction that allows the application of L'Hopital's Rule or other limit techniques.

Learn how to evaluate the limit \(\lim_{x \to 0^+} \sqrt[5]{x} \ln(x)\) step-by-step with detailed mathematical analysis. This problem involves limits, indeterminate forms, and the application of L'Hopital's Rule. Suitable for advanced high school students and beyond.

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