Question

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

$$ Answer: As t approaches π/2, the limit of the vector-valued function is: $$(-1\mathbf{i} - 4\mathbf{j} + 1\mathbf{k})$$

Step-by-step solution

Evaluate the limit for the i-component

The i-component of the given vector function is cos(2t). To find its limit as t approaches π/2, simply substitute π/2 into the function: $$\lim_{t \rightarrow \pi/2} \cos(2t) = \cos(2(\pi/2)) = \cos(\pi)$$ Since cos(π) = -1, the limit for the i-component is: $$\lim_{t \rightarrow \pi/2}\cos(2t) \mathbf{i} = -1\mathbf{i}$$

Evaluate the limit for the j-component

The j-component of the given vector function is -4sin(t). To find its limit as t approaches π/2, substitute π/2 into the function: $$\lim_{t \rightarrow \pi/2} -4\sin(t) = -4\sin(\pi/2)$$ Since sin(π/2) = 1, the limit for the j-component is: $$\lim_{t \rightarrow \pi/2} -4\sin(t) \mathbf{j} = -4\mathbf{j}$$

Evaluate the limit for the k-component

The k-component of the given vector function is (2t/π). To find its limit as t approaches π/2, substitute π/2 into the function: $$\lim_{t \rightarrow \pi/2} \frac{2t}{\pi} = \frac{2(\pi/2)}{\pi}$$ Simplifying the expression, we get: $$\lim_{t \rightarrow \pi/2} \frac{2t}{\pi} \mathbf{k} =1\mathbf{k}$$

Combine the components

Now that we have evaluated the limit for each component, we can combine them to obtain the limit of the entire vector-valued function: $$\lim_{t \rightarrow \pi/2} \left(\cos(2t) \mathbf{i} - 4\sin(t) \mathbf{j} + \frac{2t}{\pi} \mathbf{k}\right) = (-1\mathbf{i} - 4\mathbf{j} + 1\mathbf{k})$$