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Chapter 11: 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 \(4\mathbf{i}+3\mathbf{j}-\mathbf{k}\).

Step by step solution

01

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.
02

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.
03

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.
04

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: $$\lim_{t \rightarrow \ln 2}\left(2 e^{t} \mathbf{i}+6 e^{-t} \mathbf{j}-4 e^{-2 t} \mathbf{k}\right) = 4\mathbf{i}+3\mathbf{j}-\mathbf{k}$$

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Most popular questions from this chapter

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Geometric-arithmetic mean Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).

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Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and \(f\) are real numbers. It can be shown that this curve lies in a plane. Graph the following curve and describe it. $$\begin{aligned}\mathbf{r}(t)=&\left(\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{i}+\left(-\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{j} \\\&+\left(\frac{1}{\sqrt{3}} \sin t\right) \mathbf{k} \end{aligned}$$
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