To find the limit of the expression $\lim_{x \to 0^+} \sqrt[3]{x} \ln(x)$, we can break down the problem into parts.

We are evaluating:

$\lim_{x \to 0^+} \sqrt[3]{x} \ln(x)$

Step 1: Analyze the behavior of each term as $x \to 0^+$.

1. Cube Root of x:
   $\sqrt[3]{x} = x^{1/3}$. As $x \to 0^+$, $x^{1/3} \to 0$.

2. Natural Logarithm of x:
   $\ln(x)$ tends to $-\infty$ as $x \to 0^+$.

Step 2: Combining the two terms.

We have an indeterminate form $0 \cdot (-\infty)$ as $x \to 0^+$. This suggests we need to analyze it more carefully, possibly using techniques such as L'Hopital's Rule, or rewriting the expression.

Step 3: Apply L'Hopital's Rule.

We can rewrite the expression as:

$\sqrt[3]{x} \ln(x) = \frac{\ln(x)}{x^{-1/3}}$

This is now in the form $\frac{-\infty}{\infty}$, which is an indeterminate form, and we can apply L'Hopital's Rule. To do that, we differentiate the numerator and denominator with respect to $x$:

• Derivative of the numerator $\ln(x)$ is $\frac{1}{x}$.
• Derivative of the denominator $x^{-1/3}$ is $-\frac{1}{3} x^{-4/3}$.

So, applying L'Hopital's Rule:

$\lim_{x \to 0^+} \frac{\ln(x)}{x^{-1/3}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{3} x^{-4/3}} = \lim_{x \to 0^+} -3 x^{7/3}$

Step 4: Evaluate the limit.

As $x \to 0^+$, $x^{7/3} \to 0$. Therefore, the limit is:

$\lim_{x \to 0^+} -3 x^{7/3} = 0$

Final Answer:

$\lim_{x \to 0^+} \sqrt[3]{x} \ln(x) = 0$

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Would you like further explanation or have any questions? Here are some related questions:

1. What happens to $\lim_{x \to 0^+} \sqrt{x} \ln(x)$?
2. How does L'Hopital's Rule apply to other indeterminate forms like $0^0$ or $\infty/\infty$?
3. How can I apply L'Hopital's Rule to limits involving exponential functions?
4. Can you explain why $\ln(x) \to -\infty$ as $x \to 0^+$?
5. How would this limit change if the expression was $\sqrt[3]{x} \sin(x)$?

Tip: When you encounter limits of the form $0 \cdot \infty$, it's often useful to rewrite them as fractions and apply L'Hopital's Rule.