To compute the limit

$\lim_{x \to 0^+} 2 \sin(x) \ln(x),$

we’ll examine the behavior of each part of the expression as $x \to 0^+$.

1. Examine $\sin(x)$:

   • As $x \to 0^+$, $\sin(x) \approx x$. So, $\sin(x)$ approaches $0$ from the positive side.

2. Examine $\ln(x)$:

   • As $x \to 0^+$, $\ln(x) \to -\infty$.

This means we are dealing with an indeterminate form of $0 \cdot (-\infty)$, which suggests rewriting it as a fraction for L'Hôpital's Rule.

Rewriting the Expression

Rewrite $2 \sin(x) \ln(x)$ as: $2 \sin(x) \ln(x) = \frac{2 \ln(x)}{\frac{1}{\sin(x)}}.$

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

• $\ln(x) \to -\infty$,
• $\frac{1}{\sin(x)} \to \infty$.

So this is an $\frac{-\infty}{\infty}$ form, which allows us to apply L'Hôpital's Rule.

Applying L'Hôpital's Rule

Differentiate the numerator and the denominator with respect to $x$:

1. Numerator $2 \ln(x)$: $\frac{d}{dx}(2 \ln(x)) = \frac{2}{x}.$

2. Denominator $\frac{1}{\sin(x)}$: $\frac{d}{dx} \left( \frac{1}{\sin(x)} \right) = -\frac{\cos(x)}{\sin^2(x)}.$

So the limit becomes: $\lim_{x \to 0^+} \frac{\frac{2}{x}}{-\frac{\cos(x)}{\sin^2(x)}} = \lim_{x \to 0^+} \frac{-2 \sin^2(x)}{x \cos(x)}.$

Simplifying Further

Using $\sin(x) \approx x$ as $x \to 0$, we get: $\lim_{x \to 0^+} \frac{-2 x^2}{x \cdot 1} = \lim_{x \to 0^+} -2x = 0.$

Conclusion

$\lim_{x \to 0^+} 2 \sin(x) \ln(x) = 0.$

Would you like more detailed steps or have any questions?

1. Why do we approximate $\sin(x)$ by $x$ as $x \to 0$?
2. What other types of indeterminate forms can we solve using L'Hôpital's Rule?
3. Can you apply L'Hôpital's Rule multiple times if needed?
4. Why is $\ln(x) \to -\infty$ as $x \to 0^+$?
5. What is the significance of rewriting the limit in fraction form?

Tip: When using L'Hôpital's Rule, make sure each application leads closer to a solvable limit to avoid repetitive differentiation.