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

Step-by-step solution

Compute the limit of the first component

We need to evaluate the limit of the first component as t approaches ln(2): $$\lim_{t \rightarrow \ln 2}(2e^t)$$ Since the exponential function is continuous, we can simply plug in t = ln(2) to get the limit: $$=2e^{\ln 2} = 2(2) = 4$$ So the limit of the first component is 4.

Compute the limit of the second component

Now we need to evaluate the limit of the second component as t approaches ln(2): $$\lim_{t \rightarrow \ln 2}(6e^{-t})$$ Similarly, as the exponential function is continuous, we can plug in t = ln(2) to get the limit: $$=6e^{-\ln 2}=6\left(\frac{1}{e^{\ln 2}}\right)=6\left(\frac{1}{2}\right)=3$$ So the limit of the second component is 3.

Compute the limit of the third component

Finally, we need to evaluate the limit of the third component as t approaches ln(2): $$\lim_{t \rightarrow \ln 2}(-4e^{-2t})$$ Once again, the exponential function is continuous, so we can plug in t = ln(2) to get the limit: $$=-4e^{-2\ln 2}=-4\left(\frac{1}{e^{2\ln 2}}\right)=-4\left(\frac{1}{2^2}\right)=-1$$ So the limit of the third component is -1.

Combine the limits of the components

Now that we have found the limits of each component, we can combine them to form the final vector limit: $$\lim_{t \rightarrow \ln 2}\left(2 e^{t} \mathbf{i}+6 e^{-t} \mathbf{j}-4 e^{-2 t} \mathbf{k}\right) = 4\mathbf{i}+3\mathbf{j}-\mathbf{k}$$