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Chapter 11: Problem 44

Evaluate the following limits. $$\lim _{t \rightarrow 2}\left(\frac{t}{t^{2}+1} \mathbf{i}-4 e^{-t} \sin \pi t \mathbf{j}+\frac{1}{\sqrt{4 t+1}} \mathbf{k}\right)$$

Short Answer

Expert verified
Answer: The limit of the given expression as t approaches 2 is: $$\frac{2}{5}\mathbf{i} + \frac{1}{3}\mathbf{k}$$

Step by step solution

01

Identify the components

First, we will break the given limit into three components, each corresponding to a specific vector (i, j, or k). The i-component is: $$\frac{t}{t^{2}+1}$$ The j-component is: $$-4 e^{-t} \sin \pi t$$ The k-component is: $$\frac{1}{\sqrt{4 t+1}}$$ Now we will find the limit of each component as t approaches 2:
02

Find the limit of the i-component

To find the limit of the i-component as t approaches 2, we will simply substitute t with 2 in the expression. $$\lim_{t \rightarrow 2}\left(\frac{t}{t^{2}+1}\right) =\frac{2}{2^{2}+1} = \frac{2}{5}$$
03

Find the limit of the j-component

To find the limit of the j-component as t approaches 2, we will again substitute t with 2 in the expression. $$\lim_{t \rightarrow 2}\left(-4 e^{-t} \sin \pi t\right) = -4 e^{-2} \sin (2\pi) = -4 e^{-2} \cdot 0 = 0$$
04

Find the limit of the k-component

To find the limit of the k-component as t approaches 2, we will substitute t with 2 in the expression. $$\lim_{t \rightarrow 2}\left(\frac{1}{\sqrt{4t+1}}\right) = \frac{1}{\sqrt{4\cdot 2+1}} = \frac{1}{\sqrt{9}} = \frac{1}{3}$$
05

Combine the component limits

Now we can combine the limit of each component to form the overall limit as t approaches 2. $$\lim _{t \rightarrow 2}\left(\frac{t}{t^{2}+1} \mathbf{i}-4 e^{-t} \sin \pi t \mathbf{j}+\frac{1}{\sqrt{4 t+1}} \mathbf{k}\right) = \left(\frac{2}{5}\mathbf{i} + 0\mathbf{j} + \frac{1}{3}\mathbf{k}\right)$$ So the limit of the given expression as t approaches 2 is: $$\frac{2}{5}\mathbf{i} + \frac{1}{3}\mathbf{k}$$

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