The given limit problem is:

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

Step-by-step Solution:

1. Use the trigonometric identity for the difference of sines: $\sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)$ Let $A = x$ and $B = 3x$, then: $\sin x - \sin 3x = 2 \cos\left(\frac{x + 3x}{2}\right) \sin\left(\frac{x - 3x}{2}\right)$ Simplifying: $\sin x - \sin 3x = 2 \cos(2x) \sin(-x)$ Since $\sin(-x) = -\sin x$, we get: $\sin x - \sin 3x = -2 \cos(2x) \sin x$

2. Substitute this expression into the limit: $\lim_{{x \to 0}} \frac{-2 \cos(2x) \sin x}{\sin x}$ Simplifying the expression: $\lim_{{x \to 0}} -2 \cos(2x)$

3. Evaluate the limit: As $x \to 0$, $\cos(2x) \to \cos(0) = 1$. Therefore: $\lim_{{x \to 0}} -2 \cos(2x) = -2 \times 1 = -2$

Thus, the value of the limit is:

$\boxed{-2}$

Would you like further details or explanations on any step?

Here are five related questions to deepen your understanding:

1. How can the difference of two sine functions be derived using trigonometric identities?
2. What is the significance of the small angle approximation for limits involving trigonometric functions?
3. How does L'Hopital's Rule apply to limits of the form $\frac{0}{0}$?
4. How would the solution change if the sine terms in the numerator involved different multiples of $x$?
5. How can this type of limit be evaluated graphically or numerically?

Tip: When solving limits involving trigonometric functions near zero, trigonometric identities and small-angle approximations are often very useful!