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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.

Find the limit \( \lim_{\theta \to 0} \frac{\sin \theta}{\sin 2\theta} \).

Explore the step-by-step solution for finding the limit of the expression \( \frac{\sin \theta}{\sin 2\theta} \) as \( \theta \) approaches 0. This problem covers trigonometric identities, specifically the double-angle identity, and is suitable for high school students in Grades 11-12.

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