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Chapter 11: Problem 41

Evaluate the following limits. $$\lim _{t \rightarrow \pi / 2}\left(\cos 2 t \mathbf{i}-4 \sin t \mathbf{j}+\frac{2 t}{\pi} \mathbf{k}\right)$$

Short Answer

Expert verified
Answer: The limit of the vector is $$-\mathbf{i}- 4\mathbf{j} + \mathbf{k}$$.

Step by step solution

01

Break down the components

We have the given vector: $$\lim_{t \rightarrow \pi / 2}(\cos 2t \mathbf{i}-4 \sin t \mathbf{j}+\frac{2t}{\pi} \mathbf{k})$$ We'll evaluate the limit of each component separately, so let's break down the components: - x component: $$\cos 2t$$ - y component: $$-4 \sin t$$ - z component: $$\frac{2t}{\pi}$$
02

Evaluate the limit of the x component

For this component, we have $$\cos 2t$$ and we need to find $$\lim_{t \rightarrow \pi / 2}\cos 2t$$. Since \(\cos\) is a continuous function, we can simply plug in \(\pi / 2\): $$\cos(2 \cdot (\pi / 2))$$ $$=\cos(\pi)$$ $$=-1$$
03

Evaluate the limit of the y component

Now we'll evaluate $$\lim_{t \rightarrow \pi / 2}(-4 \sin t)$$. Similar to Step 2, since \(\sin\) is also continuous, we can directly plug in \(\pi / 2\): $$-4 \sin(\pi / 2)$$ $$=-4 \cdot 1$$ $$=-4$$
04

Evaluate the limit of the z component

For the last component, we have $$\frac{2t}{\pi}$$ and need to find $$\lim_{t \rightarrow \pi / 2}\frac{2t}{\pi}$$. Since this is a continuous function as well, we can directly plug in \(\pi / 2\): $$\frac{2(\pi / 2)}{\pi}$$ As we simplify this, we get: $$\frac{\pi}{\pi}$$ $$=1$$
05

Combine the limits to find the limit of the entire vector

Now that we've found the limits for each of the components, we'll combine them to find the limit of the entire vector: $$\lim_{t \rightarrow \pi / 2}(\cos 2t \mathbf{i}-4 \sin t \mathbf{j}+\frac{2t}{\pi} \mathbf{k})=(-1 \mathbf{i} - 4 \mathbf{j} + 1 \mathbf{k})$$ The limit of the vector is: $$\lim_{t \rightarrow \pi / 2}\left(\cos 2t \mathbf{i}-4 \sin t \mathbf{j}+\frac{2t}{\pi} \mathbf{k}\right) = - \mathbf{i} - 4 \mathbf{j} + \mathbf{k}$$

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