Question

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

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

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).

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: $$ \lim_{t \rightarrow \ln 2} - 4e^{-2t} $$

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.

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.

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.

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 \(4\mathbf{i} + 3\mathbf{j} - 1\mathbf{k}\).