\( 1 \leftarrow \quad \) Evaluate the following limit. \[ \lim _{t \rightarrow \pi}\left(\cos t i-2 \sin \left(-\frac{t}{2}\right) j+\frac{t}{\pi} k\right) \] \( \lim _{t \rightarrow \pi}\left(\cos t i-2 \sin \left(-\frac{t}{2}\right) j+\frac{t}{\pi} k\right)=\square \)

To evaluate the limit \[ \lim _{t \rightarrow \pi}\left(\cos t \, i - 2 \sin \left(-\frac{t}{2}\right) \, j + \frac{t}{\pi} \, k\right), \] we will analyze each component of the vector expression as \( t \) approaches \( \pi \). 1. **First component \( \cos t \)**: \[ \cos(\pi) = -1. \] 2. **Second component \( -2 \sin \left(-\frac{t}{2}\right) \)**: Substituting \( t = \pi \), \[ -2 \sin \left(-\frac{\pi}{2}\right) = -2 \cdot (-1) = 2. \] 3. **Third component \( \frac{t}{\pi} \)**: As \( t \) approaches \( \pi \), \[ \frac{\pi}{\pi} = 1. \] Now, combining these results, we get: \[ \lim_{t \rightarrow \pi}\left(\cos t \, i - 2 \sin\left(-\frac{t}{2}\right) \, j + \frac{t}{\pi} \, k\right) = -1 \, i + 2 \, j + 1 \, k. \] Therefore, the limit is \[ \boxed{-1 \, i + 2 \, j + 1 \, k}. \]