Problem 3
Question
\(3-4=\) Find the limit $$\lim _{t \rightarrow 0}\left(e^{-3 / \mathbf{i}} \mathbf{i}+\frac{t^{2}}{\sin ^{2} t} \mathbf{j}+\cos 2 t \mathbf{k}\right)$$
Step-by-Step Solution
Verified Answer
\((\cos(3) + i\sin(3))\mathbf{i} + \mathbf{j} + \mathbf{k}\)
1Step 1: Evaluate the First Component
The first component of our vector is \(e^{-3 / i}\), where \(i\) is the imaginary unit. We can rewrite this expression using Euler's formula: \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\). Thus, \(e^{-3/i} = e^{3i} = \cos(3) + i\sin(3)\). As \(t\) approaches zero, this component is simply \(\cos(3) + i\sin(3)\). Since it doesn't involve \(t\), it doesn't change with \(t\) approaching zero.
2Step 2: Simplify the Second Component
The second component is \(\frac{t^2}{\sin^2 t}\). When \(t\) approaches zero, \(\frac{t^2}{\sin^2 t}\) approaches 1 since \(\sin t \approx t\) for small \(t\), which gives \(\frac{t}{\sin t} \approx 1\). Therefore, \(\lim_{t \to 0} \frac{t^2}{\sin^2 t} = 1\). Thus, this component goes to \(\mathbf{j}\) as \(t\) approaches zero.
3Step 3: Evaluate the Third Component
The third component is \(\cos 2t\mathbf{k}\). As \(t\) approaches zero, \(\cos 2t\) approaches \(\cos 0 = 1\). Consequently, this component becomes \(\mathbf{k}\) as \(t\) approaches zero.
4Step 4: Combine the Components
Combine the limiting values of each component as \(t\) approaches zero. The first component evaluates to \(\cos(3) + i\sin(3)\mathbf{i}\), the second component becomes \(\mathbf{j}\), and the third component becomes \(\mathbf{k}\). Thus, the limit of the entire expression is: \(\lim_{t \to 0} = (\cos(3) + i\sin(3))\mathbf{i} + \mathbf{j} + \mathbf{k}\).
Key Concepts
Euler's formulaTaylor series approximationTrigonometric limits
Euler's formula
Euler's formula beautifully bridges complex numbers and trigonometry. The formula is written as \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\). In this expression, \(e^{i\theta}\) represents an exponential function involving the imaginary unit \(i\), where \(\theta\) is an angle. This results in a point on the unit circle in the complex plane.
This formula helps us transform complex exponents into more familiar trigonometric terms, as seen in our vector function limit problem. When we encounter \(e^{-3/i}\), using Euler's formula converts it into \(\cos(3) + i\sin(3)\).
Euler's formula not only shows how exponential functions can describe rotations, but it also simplifies complex calculations. By understanding how to apply this formula, you'll be able to evaluate complex exponential expressions more intuitively. This is especially useful in physics, engineering, and other fields where complex numbers often appear.
This formula helps us transform complex exponents into more familiar trigonometric terms, as seen in our vector function limit problem. When we encounter \(e^{-3/i}\), using Euler's formula converts it into \(\cos(3) + i\sin(3)\).
Euler's formula not only shows how exponential functions can describe rotations, but it also simplifies complex calculations. By understanding how to apply this formula, you'll be able to evaluate complex exponential expressions more intuitively. This is especially useful in physics, engineering, and other fields where complex numbers often appear.
Taylor series approximation
Taylor series approximation is a tool to express functions as infinite sums of their derivatives evaluated at a point. It is incredibly useful for approximating functions that are difficult to analyze otherwise. For example, the sine function can be approximated by its Taylor series:
\[ \sin(t) \approx t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots \]
When you want to evaluate limits as \(t\) approaches zero, like in our problem, Taylor series can simplify the process. For very small values of \(t\), \(\sin(t)\) can be closely approximated by \(t\). This approximation allows us to simplify complex expressions involving trigonometric functions.
In our step-by-step solution, we used this concept to show that \(\frac{t^2}{\sin^2(t)}\) simplifies to 1 as \(t\) approaches zero, by recognizing \(\sin(t)\approx t\). This demonstrates how powerful Taylor series can be in finding limits and solving calculus problems.
\[ \sin(t) \approx t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots \]
When you want to evaluate limits as \(t\) approaches zero, like in our problem, Taylor series can simplify the process. For very small values of \(t\), \(\sin(t)\) can be closely approximated by \(t\). This approximation allows us to simplify complex expressions involving trigonometric functions.
In our step-by-step solution, we used this concept to show that \(\frac{t^2}{\sin^2(t)}\) simplifies to 1 as \(t\) approaches zero, by recognizing \(\sin(t)\approx t\). This demonstrates how powerful Taylor series can be in finding limits and solving calculus problems.
Trigonometric limits
Understanding trigonometric limits is crucial for evaluating expressions involving angles as they approach particular values. A classic example is the limit \(\lim_{t \to 0} \frac{\sin(t)}{t} = 1\). This fundamental limit forms the basis of our approximation techniques and is vital in calculus.
In our original exercise, we utilized this principle to simplify the second component \(\frac{t^2}{\sin^2(t)}\). Since \(\frac{\sin(t)}{t}\) approaches 1 as \(t\) approaches zero, and thus \(\frac{t}{\sin(t)} \approx 1\), the expression ultimately simplifies to 1.
Trigonometric limits often appear when you need to resolve indeterminate forms like \(\frac{0}{0}\), a common occurrence in calculus. Being comfortable with these limits is essential for mastering calculus topics, especially those involving derivatives and integrals of trigonometric functions.
In our original exercise, we utilized this principle to simplify the second component \(\frac{t^2}{\sin^2(t)}\). Since \(\frac{\sin(t)}{t}\) approaches 1 as \(t\) approaches zero, and thus \(\frac{t}{\sin(t)} \approx 1\), the expression ultimately simplifies to 1.
Trigonometric limits often appear when you need to resolve indeterminate forms like \(\frac{0}{0}\), a common occurrence in calculus. Being comfortable with these limits is essential for mastering calculus topics, especially those involving derivatives and integrals of trigonometric functions.
Other exercises in this chapter
Problem 3
2-10 Find \(\mathbf{a} \cdot \mathbf{b}\) $$\mathbf{a}=\left\langle- 2, \frac{1}{3}\right\rangle, \quad \mathbf{b}=\langle- 5,12\rangle$$
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Which of the points \(A(-4,0,-1), B(3,1,-5),\) and \(C(2,4,6)\) is closest to the \(y z\) -plane? Which point lies in the \(x z\) -plane?
View solution Problem 4
Find the velocity, acceleration, and speed of a particle with the given position function. Sketch the path of the particle and draw the velocity and acceleratio
View solution Problem 4
\(3-4=\) Find the limit $$\lim _{t \rightarrow 1}\left(\frac{t^{2}-t}{t-1} \mathbf{i}+\sqrt{t+8} \mathbf{j}+\frac{\sin \pi t}{\ln t} \mathbf{k}\right)$$
View solution