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.