The expression provided is:

$\lim_{{x \to 0}} \frac{3x - \sin(2x)}{x \sin(x) + \ln(5x)}$

To solve this limit, let's analyze the behavior of the numerator and denominator as $x \to 0$:

Step 1: Analyze the Numerator

In the numerator, we have: $3x - \sin(2x)$ As $x \to 0$:

1. $3x \to 0$.
2. $\sin(2x) \approx 2x$ because $\sin(kx) \approx kx$ for small $x$.

So, $3x - \sin(2x) \approx 3x - 2x = x$.

Step 2: Analyze the Denominator

In the denominator, we have: $x \sin(x) + \ln(5x)$ As $x \to 0$:

1. $x \sin(x) \approx x \cdot x = x^2$, since $\sin(x) \approx x$ for small $x$.
2. $\ln(5x) \to -\infty$ as $x \to 0$ from the positive side.

Thus, the denominator tends toward $-\infty$ due to the $\ln(5x)$ term.

Conclusion

Since the numerator approaches $x$ (a finite value) and the denominator approaches $-\infty$, the entire fraction approaches $0$.

Therefore: $\lim_{{x \to 0}} \frac{3x - \sin(2x)}{x \sin(x) + \ln(5x)} = 0$

Would you like further details on any steps, or have any questions?

Here are some related questions to consider:

1. How does L'Hôpital's Rule apply to limits that approach an indeterminate form?
2. What are some common approximations for trigonometric functions near zero?
3. How does the natural logarithm function behave as its argument approaches zero from the right?
4. Can this limit be solved using series expansion methods?
5. What would happen if the denominator had a different logarithmic function, such as $\ln(x^2)$?

Tip: For limits involving $\sin(x)$ or $\ln(x)$, remember that small-angle approximations and logarithmic properties can simplify the calculations significantly.