We need to solve the following limit:

$\lim_{{x \to 0}} \frac{x + 3 \sin x}{x}$

To tackle this, let's break the expression into two parts:

$\frac{x + 3 \sin x}{x} = \frac{x}{x} + \frac{3 \sin x}{x}$

This simplifies to:

$1 + 3 \cdot \frac{\sin x}{x}$

We know that one of the standard limits is:

$\lim_{{x \to 0}} \frac{\sin x}{x} = 1$

Now, applying this result to our expression:

$\lim_{{x \to 0}} \left( 1 + 3 \cdot \frac{\sin x}{x} \right) = 1 + 3 \cdot 1 = 1 + 3 = 4$

Thus, the limit is:

$\boxed{4}$

Would you like further details or have any questions?

Here are 5 related questions to deepen understanding:

1. What is the limit of $\lim_{{x \to 0}} \frac{\sin(x)}{x^2}$?
2. How does the small-angle approximation for $\sin(x)$ help in limit problems?
3. What is L'Hopital's rule, and could it be applied to this limit?
4. What are some common limits involving trigonometric functions?
5. How would the limit change if the function were $\frac{x + 3 \sin(x)}{x^2}$?

Tip: Remember that $\lim_{{x \to 0}} \frac{\sin x}{x} = 1$ is a crucial limit often used in trigonometric limit problems.