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Chapter 12: Problem 45

Evaluate the following limits. $$\lim _{t \rightarrow 0}\left(\frac{\sin t}{t} \mathbf{i}-\frac{e^{t}-t-1}{t} \mathbf{j}+\frac{\cos t+t^{2} / 2-1}{t^{2}} \mathbf{k}\right)$$

Short Answer

Expert verified
Answer: The limit is $\mathbf{i}$.

Step by step solution

01

Find i-component limit

To find the limit of the i component, we need to evaluate the following limit: $$ \lim_{t\to0}\frac{\sin t}{t} $$ This is a well-known limit, which is equal to 1. So, the i component is: $$ \mathbf{i}(1) $$
02

Find j-component limit

To find the limit of the j component, we need to evaluate the following limit: $$ \lim_{t\to0}\frac{e^t - t - 1}{t} $$ By applying the L'H么pital's rule, we differentiate the numerator and the denominator with respect to t and get the new limit: $$ \lim_{t\to0}\frac{e^t - 1}{1} $$ Now, we plug in t=0 to get the value of the limit, which is: $$ \frac{e^0 - 1}{1} = 0 $$ So, the j component is: $$ \mathbf{j}(0) $$
03

Find k-component limit

To find the limit of the k component, we need to evaluate the following limit: $$ \lim_{t\to0}\frac{\cos t + \frac{t^2}{2} - 1}{t^2} $$ By applying the L'H么pital's rule two times, we differentiate the numerator and denominator with respect to t and get the new limits: $$ \lim_{t\to0}\frac{-\sin t + t}{2t} = \lim_{t\to0}\frac{-\cos t + 1}{2} $$ Now, we plug in t=0 to get the value of the limit, which is: $$ \frac{-\cos 0 + 1}{2} = \frac{-1+1}{2}=0 $$ So, the k component is: $$ \mathbf{k}(0) $$
04

Combine the components

Now, combine the calculated components to find the limit of the given vector expression: $$ \lim_{t\rightarrow 0}\left(\frac{\sin t}{t}\mathbf{i}-\frac{e^{t}-t-1}{t}\mathbf{j}+\frac{\cos t+t^{2}/2-1}{t^2}\mathbf{k}\right) = \mathbf{i}(1) + \mathbf{j}(0) + \mathbf{k}(0) = \boxed{\mathbf{i}} $$

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Most popular questions from this chapter

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 fare real numbers. It can be shown that this curve lies in a plane. Find a general expression for a nonzero vector orthogonal to the plane containing the curve. $$\begin{aligned} \mathbf{r}(t)=&(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j} \\\ &+(e \cos t+f \sin t) \mathbf{k} \end{aligned}$$ where \(\langle a, c, e\rangle \times\langle b, d, f\rangle \neq \mathbf{0}.\)

The points \(P, Q, R,\) and \(S,\) joined by the vectors \(\mathbf{u}, \mathbf{v}, \mathbf{w},\) and \(\mathbf{x},\) are the vertices of a quadrilateral in \(\mathrm{R}^{3}\). The four points needn't lie in \(a\) plane (see figure). Use the following steps to prove that the line segments joining the midpoints of the sides of the quadrilateral form a parallelogram. The proof does not use a coordinate system. a. Use vector addition to show that \(\mathbf{u}+\mathbf{v}=\mathbf{w}+\mathbf{x}\) b. Let \(m\) be the vector that joins the midpoints of \(P Q\) and \(Q R\) Show that \(\mathbf{m}=(\mathbf{u}+\mathbf{v}) / 2\) c. Let n be the vector that joins the midpoints of \(P S\) and \(S R\). Show that \(\mathbf{n}=(\mathbf{x}+\mathbf{w}) / 2\) d. Combine parts (a), (b), and (c) to conclude that \(\mathbf{m}=\mathbf{n}\) e. Explain why part (d) implies that the line segments joining the midpoints of the sides of the quadrilateral form a parallelogram.

Determine the equation of the line that is perpendicular to the lines \(\mathbf{r}(t)=\langle 4 t, 1+2 t, 3 t\rangle\) and \(\mathbf{R}(s)=\langle-1+s,-7+2 s,-12+3 s\rangle\) and passes through the point of intersection of the lines \(\mathbf{r}\) and \(\mathbf{R}\).

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{aligned} &\mathbf{r}(t)=\langle 1+2 t, 7-3 t, 6+t\rangle;\\\ &\mathbf{R}(s)=\langle-9+6 s, 22-9 s, 1+3 s\rangle \end{aligned}$$

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(\mathbf{u} \cdot(\mathbf{v}+\mathbf{w})=\mathbf{u} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{w}\)
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