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.