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Chapter 12: Problem 42

Evaluate the following limits. $$\lim _{t \rightarrow \ln 2}\left(2 e^{t} \mathbf{i}+6 e^{-t} \mathbf{j}-4 e^{-2 t} \mathbf{k}\right)$$

Short Answer

Expert verified
Answer: The limit of the vector function as t approaches ln(2) is \(4\mathbf{i} + 3\mathbf{j} - 1\mathbf{k}\).

Step by step solution

01

Understand the problem and main variables

The vector function given is: $$ 2 e^{t} \mathbf{i} + 6 e^{-t} \mathbf{j} - 4 e^{-2t} \mathbf{k} $$ Our aim is to find the limit of this function as \(t\) approaches ln(2).
02

Calculate the limit of each component

We will now compute the limit of each component as t approaches ln(2). i-component: $$ \lim_{t \rightarrow \ln 2} 2e^{t} $$ j-component: $$ \lim_{t \rightarrow \ln 2} 6e^{-t} $$ k-component: $$ \lim_{t \rightarrow \ln 2} - 4e^{-2t} $$
03

Evaluate the limit of the i-component

To find the limit of the i-component, substitute ln(2) into the i-component function: $$ \lim_{t \rightarrow \ln 2} 2e^{t} = 2e^{\ln 2} = 2(2) = 4 $$ So, the limit of the i-component is 4i.
04

Evaluate the limit of the j-component

To find the limit of the j-component, substitute ln(2) into the j-component function: $$ \lim_{t \rightarrow \ln 2} 6e^{-t} = 6e^{-\ln 2} = \frac{6}{2} = 3 $$ So, the limit of the j-component is 3j.
05

Evaluate the limit of the k-component

To find the limit of the k-component, substitute ln(2) into the k-component function: $$ \lim_{t \rightarrow \ln 2} -4e^{-2t} = -4e^{-2\ln 2 } = -4\left(\frac{1}{2^2}\right) = -1 $$ So, the limit of the k-component is -1k.
06

Combine the limit of the components

Now, we combine the limits of all components to find the limit of the entire vector function: $$ \lim _{t \rightarrow \ln 2}\left(2 e^{t} \mathbf{i}+6 e^{-t} \mathbf{j}-4 e^{-2 t} \mathbf{k}\right) = 4\mathbf{i} + 3\mathbf{j} - 1\mathbf{k} $$ The limit of the given vector function as t approaches ln(2) is \(4\mathbf{i} + 3\mathbf{j} - 1\mathbf{k}\).

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