The limit expression you provided is:

$\lim_{\theta \to 0} \frac{\sin \theta}{\sin 2\theta}$

To solve this limit, let's break it down:

Step 1: Use the double-angle identity

The double-angle identity for sine is:

$\sin 2\theta = 2\sin \theta \cos \theta$

Substituting this into the limit expression:

$\lim_{\theta \to 0} \frac{\sin \theta}{2\sin \theta \cos \theta}$

Step 2: Simplify the expression

We can cancel out $\sin \theta$ from the numerator and denominator (assuming $\theta \neq 0$):

$\lim_{\theta \to 0} \frac{1}{2 \cos \theta}$

Step 3: Evaluate the limit

As $\theta$ approaches 0, $\cos \theta$ approaches $\cos 0 = 1$. Therefore:

$\lim_{\theta \to 0} \frac{1}{2 \cos \theta} = \frac{1}{2 \cdot 1} = \frac{1}{2}$

Final Answer

$\lim_{\theta \to 0} \frac{\sin \theta}{\sin 2\theta} = \frac{1}{2}$

Would you like further details or have any questions?

Here are some related questions to consider:

1. What are the other double-angle identities for trigonometric functions?
2. How would the limit change if the expression were $\frac{\sin 2\theta}{\sin \theta}$?
3. What happens to the limit if we approach a different value for $\theta$?
4. Can you generalize this limit for $\lim_{\theta \to 0} \frac{\sin \theta}{\sin n\theta}$ where $n$ is any positive integer?
5. How do limits involving trigonometric functions behave differently compared to polynomial functions?

Tip: When solving limits, always check if the expression can be simplified using trigonometric identities, which often make evaluation straightforward.