Problem 6 - Calculus Early Transcendentals

Question

\(3-6\) Find the limit. $$ \lim _{t \rightarrow \infty}\left\langle\arctan t, e^{-2 t}, \frac{\ln t}{t}\right\rangle $$

Step-by-Step Solution

Answer

The limit is \( \left\langle \frac{\pi}{2}, 0, 0 \right\rangle \)."

Step 1: Understanding the Limit

We need to find the limit of a vector as its components approach some value. The vector given is \( \langle \arctan t, e^{-2t}, \frac{\ln t}{t} \rangle \) as \( t \to \infty \). This means we need to compute the limits separately for each component.

Step 1: Compute Limit of \(\arctan t\)

As \( t \to \infty \), the function \( \arctan t \) approaches its horizontal asymptote. The limits to remember are that \( \arctan t \) approaches \( \frac{\pi}{2} \) as \( t \to \infty \). Therefore, \( \lim_{t \to \infty} \arctan t = \frac{\pi}{2} \).

Step 2: Compute Limit of \(e^{-2t}\)

As \( t \to \infty \), the exponential function \( e^{-2t} \) which is \( \frac{1}{e^{2t}} \) decreases rapidly to 0 because the exponent is negative and grows without bound. Thus, \( \lim_{t \to \infty} e^{-2t} = 0 \).

Step 3: Compute Limit of \(\frac{\ln t}{t}\)

Apply L'Hôpital's Rule since both \( \ln t \to \infty \) and \( t \to \infty \) as \( t \to \infty \). Differentiate the numerator and the denominator: \( (\ln t)' = \frac{1}{t} \) and \( t' = 1 \). So the limit becomes \( \lim_{t \to \infty} \frac{1/t}{1} = \lim_{t \to \infty} \frac{1}{t} = 0 \).

Step 5: Assemble the Vector Limit

Combine the limits of the individual components: \( \langle \frac{\pi}{2}, 0, 0 \rangle \). This is the final result for the limit of the vector as \( t \to \infty \).

Key Concepts

Vector Calculus

In vector calculus, we often explore functions where the output is a vector rather than a single number. For example, the problem here deals with the vector \( \langle \arctan t, e^{-2t}, \frac{\ln t}{t} \rangle \).
The limit of a vector function is determined by finding the limit of each individual component separately as a common parameter, in this case \( t \), approaches a specific value. This approach simplifies complex problems by breaking them down into manageable parts.
For this vector, we're evaluating the limit as \( t \to \infty \). Each function or expression in a vector describes a different behavior as \( t \) increases, and understanding these components separately can shed light on overall changes.
This step-by-step breakdown is crucial for handling higher-dimensional spaces often analyzed in vector calculus, as it helps manage complexity and provide clear insights.

L'Hôpital's Rule

The famous L'Hôpital's Rule is a technique in calculus for finding limits of indeterminate forms. When you encounter expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), this rule can become handy. In our problem, for the component \( \frac{\ln t}{t} \), both \( \ln t \) and \( t \) become infinite as \( t \rightarrow \infty \), creating a form that cannot be solved directly.
L'Hôpital's Rule allows us to differentiate both the numerator and the denominator and then evaluate the limit again.

• Differentiating \( \ln t \) gives \( \frac{1}{t} \).
• Differentiating \( t \) gives \( 1 \).

Applying these derivatives transforms the expression to \( \frac{1/t}{1} \), simplifying the evaluation of the limit to \( \lim_{t \to \infty} \frac{1}{t} = 0 \). This illustrates the effectiveness of L'Hôpital's Rule in solving limits that might initially seem complex or unsolvable, providing a systematic method to tackle such challenges.

Asymptotic Behavior

Asymptotic behavior in calculus describes how functions behave as inputs grow very large or very small. This concept is crucial for understanding limits, particularly in assessing where functions stabilize or diverge.
In our example, the function \( \arctan t \) approaches a horizontal asymptote of \( \frac{\pi}{2} \) as \( t \to \infty \). This indicates that \( \arctan t \) levels off and very slowly approaches but never actually reaches \( \frac{\pi}{2} \).
The use of asymptotic behavior helps predict the long-term trend of functions and their graph. It helps retain an insightful approach to evaluating eventual behavior at infinity, which is often where most practical applications draw meaningful insights from mathematical models.

Exponential Functions

Exponential functions have a base raised to the power of a variable, portraying very rapid growth or decay. In this exercise, the function \( e^{-2t} \) exemplifies exponential decay as \( t \to \infty \).
An important property of exponential decay is that as the exponent becomes more negative, the value approaches 0 very rapidly.

• In our example, \( e^{-2t} \) represents \( \frac{1}{e^{2t}} \), illustrating how the function trends quickly towards 0 as the denominator grows exponentially large.

Understanding these exponential functions is crucial in many real-world scenarios, such as radioactive decay or cooling processes, where the quick changes can affect significant outcomes. Recognizing the behavior of exponential functions is essential for accurate predictions and computations in various fields of science and engineering.