The limit presented is:

$\lim_{x \to 0^+} 8 \sin(x) \ln(x)$

Step-by-step solution:

1. Simplifying the expression: The given expression is $8 \sin(x) \ln(x)$. The $8$ is a constant, so we focus on the behavior of $\sin(x) \ln(x)$.

2. Behavior of the terms as $x \to 0^+$:

   • $\sin(x)$ behaves as $x$ for small $x$, since $\sin(x) \approx x$ when $x \to 0$.
   • $\ln(x)$ tends to $-\infty$ as $x \to 0^+$.

   Therefore, the product $\sin(x) \ln(x)$ becomes $x \ln(x)$.

3. Analyzing the product $x \ln(x)$: We now focus on finding the limit:

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

   This is an indeterminate form of type $0 \times (-\infty)$, so we can rewrite it as a fraction to apply L'Hopital's Rule:

   $x \ln(x) = \frac{\ln(x)}{1/x}$

4. Applying L'Hopital's Rule: To resolve this indeterminate form $\frac{-\infty}{\infty}$, we differentiate the numerator and the denominator:

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

   So we now have:

   $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0$

5. Conclusion: Since $\lim_{x \to 0^+} x \ln(x) = 0$, multiplying by 8 gives:

   $\lim_{x \to 0^+} 8 \sin(x) \ln(x) = 0$

Thus, the value of the limit is:

$\boxed{0}$

Would you like further details or have any questions?

Related Questions:

1. How is L'Hopital's Rule applied to more complex indeterminate forms?
2. What happens to the limit if we replace $\sin(x)$ with another function like $\cos(x)$?
3. How does the behavior of logarithmic functions near 0 impact limits?
4. Can this limit be solved using series expansion for $\sin(x)$?
5. What are other common cases where L'Hopital's Rule is used?

Tip:

When dealing with limits involving logarithms, rewriting the expression in a fractional form often helps apply L'Hopital's Rule efficiently.