Questions: lim as t approaches π of (cos(t)i - 9sin(-t/2)j + t/πk)

Transcript text: \[ \lim _{t \rightarrow \pi}\left(\cos t i-9 \sin \left(-\frac{t}{2}\right) j+\frac{t}{\pi} k\right) \]

Solution

Solution Steps

To evaluate the given limit, we need to find the limit of each component of the vector as \( t \) approaches \( \pi \). The vector is composed of three components: \( \cos t \), \( -9 \sin \left(-\frac{t}{2}\right) \), and \( \frac{t}{\pi} \). We will evaluate each of these limits separately and then combine them to get the final result.

Evaluate the limit of \( \cos t \) as \( t \rightarrow \pi \).
Evaluate the limit of \( -9 \sin \left(-\frac{t}{2}\right) \) as \( t \rightarrow \pi \).
Evaluate the limit of \( \frac{t}{\pi} \) as \( t \rightarrow \pi \).
Combine the results to get the final vector limit.

Step 1: Evaluate the Limit of \(\cos t\) as \(t \rightarrow \pi\)

The first component of the vector is \(\cos t\). As \(t\) approaches \(\pi\), we have: \[ \lim_{t \rightarrow \pi} \cos t = \cos(\pi) = -1 \]

Step 2: Evaluate the Limit of \(-9 \sin \left(-\frac{t}{2}\right)\) as \(t \rightarrow \pi\)

The second component of the vector is \(-9 \sin \left(-\frac{t}{2}\right)\). As \(t\) approaches \(\pi\), we have: \[ \lim_{t \rightarrow \pi} -9 \sin \left(-\frac{t}{2}\right) = -9 \sin \left(-\frac{\pi}{2}\right) = -9 \times (-1) = 9 \]

Step 3: Evaluate the Limit of \(\frac{t}{\pi}\) as \(t \rightarrow \pi\)

The third component of the vector is \(\frac{t}{\pi}\). As \(t\) approaches \(\pi\), we have: \[ \lim_{t \rightarrow \pi} \frac{t}{\pi} = \frac{\pi}{\pi} = 1 \]