Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{t \rightarrow \infty}\left[\frac{t \sin t}{t^{2}} \mathbf{i}-\frac{7 t^{3}}{t^{3}-3 t} \mathbf{j}-\frac{\sin t}{t} \mathbf{k}\right] $$

Step-by-Step Solution

Answer

The limit is \(-7 \mathbf{j}\).

Step 1: Analyze Each Component Separately

The given vector function is \( \frac{t \sin t}{t^2} \mathbf{i} - \frac{7t^3}{t^3 - 3t} \mathbf{j} - \frac{\sin t}{t} \mathbf{k} \). We need to find the limit as \( t \to \infty \). We'll handle each component of the vector separately.

Step 2: Limit of the i-component

Examine the first component: \( \frac{t \sin t}{t^2} \). We can simplify it as \( \frac{\sin t}{t} \) because \( \frac{t}{t^2} = \frac{1}{t} \). As \( t \to \infty \), \( \frac{\sin t}{t} \to 0 \) since \( \sin t \) is bounded between -1 and 1 and divides by an ever-increasing \( t \). Therefore, \( \lim_{t \to \infty} \frac{t \sin t}{t^2} = 0 \).

Step 3: Limit of the j-component

Now consider the second component: \( \frac{7t^3}{t^3 - 3t} \). Simplify this to \( \frac{7}{1 - \frac{3}{t^2}} \) by dividing the numerator and denominator by \( t^3 \). As \( t \to \infty \), the term \( \frac{3}{t^2} \to 0 \). Hence, \( \lim_{t \to \infty} \frac{7t^3}{t^3 - 3t} = 7 \).

Step 4: Limit of the k-component

For the third component \( \frac{\sin t}{t} \), as \( t \to \infty \), it approaches \( 0 \) since \( \sin t \) is bounded and \( t \to \infty \) causes \( \frac{1}{t} \to 0 \). Thus, \( \lim_{t \to \infty} \frac{\sin t}{t} = 0 \).

Step 5: Combine Results

Combine the limits of each component from Steps 2, 3, and 4:\[ \lim_{t \to \infty} \left[ \frac{t \sin t}{t^2} \mathbf{i} - \frac{7t^3}{t^3 - 3t} \mathbf{j} - \frac{\sin t}{t} \mathbf{k} \right] = 0 \mathbf{i} - 7 \mathbf{j} - 0 \mathbf{k} = -7 \mathbf{j}. \]

Key Concepts

Vector Calculus

Vector Calculus is a branch of mathematics that deals with vectors and functions of vector variables. In the context of finding limits of vector functions, each component of the vector is treated as a separate scalar function. This means that to find the overall limit of a vector function, we handle each part individually. This technique makes understanding complex vector limits simpler, as we can focus on one dimension at a time.

In our exercise, the vector function is given in terms of the standard basis vectors:

• \( rac{t \sin t}{t^2} \mathbf{i} \) is the first component.
• \( -\frac{7t^3}{t^3 - 3t} \mathbf{j} \) is the second component.
• \( -\frac{\sin t}{t} \mathbf{k} \) is the third component.

Each of these components behaves separately as we approach the limit. The principles of vector calculus allow us to analyze them independently and then combine the results.

Limit Laws

Limit laws are essential tools in calculus that guide us in finding the limits of functions. They provide rules which help simplify and calculate limits effectively. These include familiar laws like direct substitution, limit of a sum, limit of a product, and more.

For our exercise, the limit laws allow us to:

• Simplify each vector component individually.
• Determine the limit of each simplified component.
• Combine these limits to find the overall limit of the vector function.

Applying limit laws involves checking whether we can directly substitute the limit value or if algebraic manipulation is needed first. In each component, algebraic simplifications made the application of limits straightforward. The limit of the sum or difference is simply the sum or difference of the limits of the components.

Component-wise Limits

Component-wise Limits refer to the process of finding the limit of each individual element of a vector function separately. This method is effective and often necessary due to the nature of vector functions.

Consider our task:

• The \(\mathbf{i}\) component simplifies as: \(\frac{\sin t}{t}\) which tends to 0 as \( t \to \infty \).
• The \(\mathbf{j}\) component simplifies by factoring: \(\frac{7}{1 - \frac{3}{t^2}}\) where only the numerator contributes significantly as \( t \to \infty \), leading to 7.
• The \(\mathbf{k}\) component: \(\frac{\sin t}{t}\) also tends to 0 as \( t \to \infty \).

Using component-wise limits helps manage complex expressions by breaking them down into simpler, more manageable parts. It ensures each component's behavior is accounted for precisely as the variable approaches infinity.

Transcendental Functions

Transcendental functions, such as trigonometric and exponential functions, play a unique role in calculus. They do not follow polynomial function rules and often have their own unique properties.

In our example, the function \( \sin t \) is a transcendental function. Its property of being bounded—meaning it always stays between -1 and 1—was crucial in determining the limits of components involving \( \sin t \).

• For the \(\mathbf{i}\) and \(\mathbf{k}\) components, \( \sin t \) was used in the fraction \( \frac{\sin t}{t} \), which tends towards zero due to this bounded behavior.
• Understanding how transcendental functions behave, particularly as contextually important as \( t \to \infty \), allows for a precise calculation of limits.

These functions often necessitate special consideration, usually through approximations or leveraging their properties, to ensure proper limit evaluation.