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Chapter 12: 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 vector-valued function as t approaches 2 is $\frac{2}{5} \mathbf{i} + \frac{1}{3} \mathbf{k}$.

Step by step solution

01

Evaluate the i-component of the limit

To find the limit of the i-component, we need to evaluate the following limit: $$\lim_{t \rightarrow 2} \frac{t}{t^2 + 1}$$ If we substitute t = 2 in the expression, we get: $$\frac{2}{2^2 + 1} = \frac{2}{5}$$ So, the limit of the i-component is: $$\lim_{t \rightarrow 2} \frac{t}{t^2 + 1} = \frac{2}{5} \mathbf{i}$$
02

Evaluate the j-component of the limit

To find the limit of the j-component, we need to evaluate the following limit: $$\lim_{t \rightarrow 2} -4 e^{-t} \sin \pi t$$ We can use the properties of limits and limits involving exponential and trigonometric functions to compute this limit: $$\lim_{t \rightarrow 2} -4 e^{-t} \sin \pi t = -4e^{-2} \sin (2\pi) = -4e^{-2}(0) = 0 \mathbf{j}$$
03

Evaluate the k-component of the limit

To find the limit of the k-component, we need to evaluate the following limit: $$\lim_{t \rightarrow 2} \frac{1}{\sqrt{4t+1}}$$ If we substitute t = 2 in the expression, we get: $$\frac{1}{\sqrt{4(2)+1}} = \frac{1}{\sqrt{9}} = \frac{1}{3}$$ So, the limit of the k-component is: $$\lim_{t \rightarrow 2} \frac{1}{\sqrt{4t+1}} = \frac{1}{3} \mathbf{k}$$
04

Combine the component limits

Now that we have found the limits of each component, we can combine them to find the limit of the entire vector-valued function: $$\lim _{t \rightarrow 2}\left(\frac{t}{t^{2}+1} \mathbf{i}-4 e^{-t} \sin \pi t \mathbf{j}+\frac{1}{\sqrt{4 t+1}} \mathbf{k}\right) = \frac{2}{5} \mathbf{i} + 0 \mathbf{j} + \frac{1}{3} \mathbf{k}$$ So, the limit of the given expression is: $$\boxed{\frac{2}{5} \mathbf{i} + \frac{1}{3} \mathbf{k}}$$

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Most popular questions from this chapter

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