To solve this limit:

$\lim_{x \to 0^+} 3 \sin(x) \ln(x),$

we need to examine the behavior of each component as $x \to 0^+$.

Step-by-Step Solution

1. Analyze the components of the expression:

   • As $x \to 0^+$, $\sin(x) \approx x$, because $\sin(x)$ is close to $x$ for small values of $x$.
   • Therefore, $3 \sin(x) \ln(x) \approx 3 x \ln(x)$ as $x \to 0^+$.

2. Rewrite the expression:

   • The limit can now be expressed as: $\lim_{x \to 0^+} 3 x \ln(x).$

3. Identify the indeterminate form:

   • As $x \to 0^+$, $\ln(x) \to -\infty$ and $x \to 0^+$. Thus, $x \ln(x)$ is of the indeterminate form $0 \cdot (-\infty)$.

4. Rewrite using L'Hôpital's Rule:

   • To apply L'Hôpital's Rule, rewrite $x \ln(x)$ as $\frac{\ln(x)}{\frac{1}{x}}$.
   • Then, the limit becomes: $\lim_{x \to 0^+} 3 \frac{\ln(x)}{\frac{1}{x}}.$
   • This is now in the form $\frac{-\infty}{\infty}$, which is suitable for L'Hôpital's Rule.

5. Apply L'Hôpital's Rule:

   • Differentiate the numerator and the denominator with respect to $x$:
     • The derivative of $\ln(x)$ is $\frac{1}{x}$.
     • The derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$.
   • Thus, we have: $\lim_{x \to 0^+} 3 \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} 3 \cdot -x = \lim_{x \to 0^+} -3x.$

6. Evaluate the final limit:

   • As $x \to 0^+$, $-3x \to 0$.
   • Therefore, $\lim_{x \to 0^+} 3 \sin(x) \ln(x) = 0.$

Final Answer

$\lim_{x \to 0^+} 3 \sin(x) \ln(x) = 0.$

Would you like more details on any step, or do you have additional questions?

Here are five related questions to expand your understanding:

1. What is the purpose of using L'Hôpital's Rule in solving limits?
2. Why is $\sin(x) \approx x$ a valid approximation as $x \to 0$?
3. How can we recognize an indeterminate form in limits?
4. What other methods could we use to evaluate limits besides L'Hôpital's Rule?
5. How does L'Hôpital's Rule apply to cases involving infinity?

Tip: When applying L'Hôpital's Rule, make sure to check that the limit is indeed an indeterminate form (such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$) before proceeding with differentiation.