Problem 11 - Thomas Calculus

Question

Find the limits in Exercises \(9-12\) . $$\lim _{t \rightarrow-\infty} \frac{2-t+\sin t}{t+\cos t}$$

Step-by-Step Solution

Answer

The limit is \(-1\).

Step 1: Analyze the Expression

The given limit is \( \lim _{t \rightarrow -\infty} \frac{2-t+\sin t}{t+\cos t} \). To find this limit, we need to analyze the behavior of both the numerator and the denominator as \( t \to -\infty \).

Step 2: Simplify the Dominant Terms

Both in the numerator \( 2 - t + \sin t \) and in the denominator \( t + \cos t \), the terms \(-t\) and \(t\) dominate because as \(t\) approaches \(-\infty\), terms not involving \(t\) (\(2, \sin t, \cos t\)) remain bounded and become negligible.

Step 3: Factor out \(t\) from the Expression

Factor \( t \) out of both the numerator and the denominator. The expression becomes \( \frac{t(-1 + \frac{2}{t} + \frac{\sin t}{t})}{t(1 + \frac{\cos t}{t})} \), which simplifies to \( \frac{-1 + \frac{2}{t} + \frac{\sin t}{t}}{1 + \frac{\cos t}{t}} \).

Step 4: Take the Limit as \( t \to -\infty \)

Since \( \frac{2}{t} \to 0 \), \( \frac{\sin t}{t} \to 0 \), and \( \frac{\cos t}{t} \to 0 \) as \( t \to -\infty \), the expression simplifies to \( \frac{-1}{1} = -1 \).

Step 5: Conclude the Limit

The limit is \( \lim _{t \rightarrow -\infty} \frac{2-t+\sin t}{t+\cos t} = -1 \).

Key Concepts

Asymptotic Analysis

In calculus, asymptotic analysis helps us understand the behavior of functions as the input approaches large values. When dealing with the problem \( \lim _{t \rightarrow -\infty} \frac{2-t+\sin t}{t+\cos t} \), we use asymptotic analysis to simplify the expression. The goal here is to find out which terms in the numerator and denominator have the most influence as \( t \) tends to \(-\infty\).
To achieve this, we look at the term behavior individually:

• Terms like \(2\), \(\sin t\), and \(\cos t\) are bounded, meaning they don't grow larger or smaller as \( t \to -\infty \).
• Terms that include \(t\), such as \(-t\) and \(t\), do become quite large, making them dominant.

Asymptotic analysis helps us disregard negligible terms like \(2\), \(\sin t\), and \(\cos t\), simplifying our work focused on the dominant terms for limits. This makes it easier to calculate limits at infinity.

Dominant Terms

When solving for limits, recognizing dominant terms is crucial. In our exercise, the expression \(\frac{2-t+\sin t}{t+\cos t}\) can be simplified by focusing on the dominant terms. These are the terms that have the biggest impact on the limit's value when \( t \) approaches \(-\infty\).
In both the numerator \(2-t+\sin t\) and the denominator \(t+\cos t\), it's evident that the terms involving \(t\) dominate:

• The \(-t\) in the numerator.
• The \(t\) in the denominator.

Because \(2\), \(\sin t\), and \(\cos t\) are bounded, we can consider them negligible at \( t \to -\infty \). Therefore, our dominant terms inference allows us to simplify the expression by factoring out \(t\) and reducing the complexity of evaluating limits.

Limits at Infinity

Understanding limits at infinity is a key concept in calculus, especially when dealing with expressions like \(\frac{2-t+\sin t}{t+\cos t}\). This type of limit looks at what happens to our expression as \( t \) moves toward \(-\infty\).
In our solution, after factoring out \(t\) and eliminating less significant terms, the expression becomes \(\frac{-1 + \frac{2}{t} + \frac{\sin t}{t}}{1 + \frac{\cos t}{t}}\). As \( t \to -\infty \), fractions like \(\frac{2}{t}\), \(\frac{\sin t}{t}\), and \(\frac{\cos t}{t}\) all converge to zero because the numerator remains constant while the denominator \(t\) grows indefinitely.

• Therefore, \(\frac{2}{t} \to 0\), \(\frac{\sin t}{t} \to 0\), and \(\frac{\cos t}{t} \to 0\).

Thus, the limit simplifies to \(\frac{-1}{1} = -1\). Understanding limits at infinity allows us to understand how expressions "settle" into values as inputs grow larger and larger.