Evaluate the following limits :
`Lim_(xto pi) (sin 2x)/(sinx)`

Answer

To evaluate the limit \( \lim_{x \to \pi} \frac{\sin 2x}{\sin x} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the limit expression**:
We need to evaluate \( \lim_{x \to \pi} \frac{\sin 2x}{\sin x} \).

2. **Use the double angle identity for sine**:
Recall the identity \( \sin 2x = 2 \sin x \cos x \). We can substitute this into our limit:
\[ \lim_{x \to \pi} \frac{\sin 2x}{\sin x} = \lim_{x \to \pi} \frac{2 \sin x \cos x}{\sin x} \]

3. **Simplify the expression**:
We can cancel \( \sin x \) in the numerator and denominator (as long as \( \sin x \neq 0 \)):
\[ \lim_{x \to \pi} 2 \cos x \]

4. **Evaluate the limit**:
Now we can directly substitute \( x = \pi \) into the simplified expression:
\[ 2 \cos(\pi) = 2 \times (-1) = -2 \]

5. **Conclusion**:
Therefore, the limit is:
\[ \lim_{x \to \pi} \frac{\sin 2x}{\sin x} = -2 \]