Evaluate the following :
`lim_(theta to 0)(1-costheta)/(1-cos3theta)`

Answer

To evaluate the limit

\[ \lim_{\theta \to 0} \frac{1 - \cos \theta}{1 - \cos 3\theta}, \]

we start by substituting \(\theta = 0\):

1. **Substitute \(\theta = 0\)**:
\[ \frac{1 - \cos(0)}{1 - \cos(3 \cdot 0)} = \frac{1 - 1}{1 - 1} = \frac{0}{0}. \]
This gives us an indeterminate form \( \frac{0}{0} \).

**Hint**: When you encounter an indeterminate form, consider using L'Hôpital's Rule.

2. **Apply L'Hôpital's Rule**:
Since we have \( \frac{0}{0} \), we can differentiate the numerator and the denominator with respect to \(\theta\).

- Differentiate the numerator:
\[ \frac{d}{d\theta}(1 - \cos \theta) = \sin \theta. \]
- Differentiate the denominator:
\[ \frac{d}{d\theta}(1 - \cos 3\theta) = 3\sin(3\theta). \]

3. **Rewrite the limit**:
Now we can rewrite the limit using the derivatives:
\[ \lim_{\theta \to 0} \frac{\sin \theta}{3 \sin(3\theta)}. \]

**Hint**: Simplifying the limit can help you evaluate it more easily.

4. **Substitute \(\theta = 0\) again**:
Substitute \(\theta = 0\):
\[ \frac{\sin(0)}{3 \sin(3 \cdot 0)} = \frac{0}{3 \cdot 0} = \frac{0}{0}. \]
We still have an indeterminate form.

5. **Apply L'Hôpital's Rule again**:
Differentiate the numerator and denominator again:
- Differentiate the numerator:
\[ \frac{d}{d\theta}(\sin \theta) = \cos \theta. \]
- Differentiate the denominator:
\[ \frac{d}{d\theta}(3 \sin(3\theta)) = 9 \cos(3\theta). \]

6. **Rewrite the limit again**:
Now we can rewrite the limit:
\[ \lim_{\theta \to 0} \frac{\cos \theta}{9 \cos(3\theta)}. \]

7. **Substitute \(\theta = 0\) one last time**:
Substitute \(\theta = 0\):
\[ \frac{\cos(0)}{9 \cos(3 \cdot 0)} = \frac{1}{9 \cdot 1} = \frac{1}{9}. \]

Thus, the final answer is

\[ \boxed{\frac{1}{9}}. \]