Evaluate the following :
`lim_(theta to 0)(sinatheta)/(sinbtheta)`

Answer

To evaluate the limit \( \lim_{\theta \to 0} \frac{\sin(a\theta)}{\sin(b\theta)} \), we can use the standard limit property that states:

\[ \lim_{x \to 0} \frac{\sin(kx)}{kx} = 1 \]

for any constant \( k \).

### Step-by-Step Solution:

1. **Rewrite the Limit**:
We start with the limit:
\[ L = \lim_{\theta \to 0} \frac{\sin(a\theta)}{\sin(b\theta)} \]

2. **Multiply and Divide by \( a\theta \) and \( b\theta \)**:
To utilize the known limit property, we can rewrite the limit as follows:
\[ L = \lim_{\theta \to 0} \frac{\sin(a\theta)}{a\theta} \cdot \frac{a\theta}{b\theta} \cdot \frac{b\theta}{\sin(b\theta)} \]
This gives us:
\[ L = \lim_{\theta \to 0} \left( \frac{\sin(a\theta)}{a\theta} \cdot \frac{a}{b} \cdot \frac{b\theta}{\sin(b\theta)} \right) \]

3. **Apply the Limit**:
Now we can apply the limit:
- As \( \theta \to 0 \), \( \frac{\sin(a\theta)}{a\theta} \to 1 \)
- As \( \theta \to 0 \), \( \frac{b\theta}{\sin(b\theta)} \to 1 \)

Therefore, we have:
\[ L = 1 \cdot \frac{a}{b} \cdot 1 = \frac{a}{b} \]

4. **Final Result**:
Thus, the limit evaluates to:
\[ \lim_{\theta \to 0} \frac{\sin(a\theta)}{\sin(b\theta)} = \frac{a}{b} \]