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The given expression is:

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

To evaluate this limit, we can apply the small-angle approximation:

\[
\lim_{{x \to 0}} \frac{\sin(kx)}{kx} = 1
\]

Using this, let's manipulate the given expression. First, rewrite the limit in a more convenient form:

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

Now, introduce the factor \( 5x \) inside the sine term by multiplying and dividing by 5:

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

Since \(\lim_{{x \to 0}} \frac{5x}{\sin(5x)} = 1\), the expression simplifies to:

\[
= \frac{3}{5} \cdot 1 = \frac{3}{5}
\]

Thus, the value of the limit is:

\[
\boxed{\frac{3}{5}}
\]

Would you like further clarification or details on this solution?

### Related Questions:
1. How does L'Hopital's rule work in this case?
2. Can the small-angle approximation be used for other trigonometric functions?
3. What is the behavior of \(\sin(kx)\) as \(x \to 0\) for different values of \(k\)?
4. How would the limit change if the denominator was \(\sin(3x)\) instead of \(\sin(5x)\)?
5. What happens if we don't use the small-angle approximation and apply a different method?

### Tip:
For limits involving trigonometric functions near zero, the small-angle approximation is a very handy tool.

Evaluate the limit: \(\lim_{{x \to 0}} \frac{3x}{\sin(5x)}\)

Learn how to evaluate the limit \(\lim_{{x \to 0}} \frac{3x}{\sin(5x)}\) using the small-angle approximation. This problem covers key concepts in trigonometric limits and is suitable for students in Grades 11-12.

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