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The expression given is:

\[
\lim_{x \to 0} \frac{\sin 3\theta}{5\theta}
\]

To solve this limit, let’s first rewrite it in a more convenient form.

### Step 1: Factor out constants
Notice that the argument inside the sine function is \(3\theta\), while the denominator has \(5\theta\). We can factor out constants to get:

\[
\lim_{\theta \to 0} \frac{\sin(3\theta)}{5\theta} = \frac{3}{5} \lim_{\theta \to 0} \frac{\sin(3\theta)}{3\theta}
\]

### Step 2: Use the standard limit property
The standard limit \(\lim_{\theta \to 0} \frac{\sin(k\theta)}{k\theta} = 1\) for any constant \(k\) can be applied here. Hence:

\[
\frac{3}{5} \lim_{\theta \to 0} \frac{\sin(3\theta)}{3\theta} = \frac{3}{5} \cdot 1 = \frac{3}{5}
\]

### Final Answer
Thus, the limit is:

\[
\lim_{\theta \to 0} \frac{\sin(3\theta)}{5\theta} = \frac{3}{5}
\]

Would you like further details or have any other questions?

### Related Questions:
1. How would the solution change if the limit was \(\lim_{\theta \to 0} \frac{\sin(2\theta)}{4\theta}\)?
2. What is the significance of the standard limit \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) in calculus?
3. How do you solve limits involving trigonometric functions when the argument inside the sine function is not linear?
4. What happens to the limit if you substitute \(\theta\) with another function of \(x\)?
5. Can you apply L'Hôpital's Rule to solve this limit? Why or why not?

### Tip:
When dealing with trigonometric limits where the argument of sine is a multiple of the variable, always consider factoring out constants and applying the standard limit formula to simplify the expression.

Learn how to solve the limit lim_{\theta \to 0} \frac{\sin 3\theta}{5\theta} with a detailed step-by-step solution. This problem covers limits involving trigonometric functions and applies the standard limit property effectively. Suitable for advanced high school students and undergraduate level learners.

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