Problem 125 - Thomas Calculus

Question

In Exercises \(125-126,\) use the definition of convergence to prove the given limit. $$\lim _{n \rightarrow \infty} \frac{\sin n}{n}=0$$

Step-by-Step Solution

Answer

The limit is 0 since \( \lim_{n \to \infty} \frac{\sin n}{n} = 0 \) for every \( \epsilon > 0 \).

Step 1: Understand the Limit Definition of Convergence

In this exercise, we need to use the definition of convergence to prove that \( \lim_{n \to \infty} \frac{\sin n}{n} = 0 \). According to the definition, for any \( \epsilon > 0 \), there exists a positive integer \( N \) such that for all \( n > N \), the inequality \( \left| \frac{\sin n}{n} - 0 \right| < \epsilon \) holds true.

Step 2: Manipulate the Expression

Since \( -1 \leq \sin n \leq 1 \) for any integer \( n \), it follows that \( -\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n} \). Thus, \( \left| \frac{\sin n}{n} \right| \leq \frac{1}{n} \).

Step 3: Choose an Appropriate N

Given that \( \left| \frac{\sin n}{n} \right| \leq \frac{1}{n} \), we can ensure \( \frac{1}{n} < \epsilon \) by choosing \( n > \frac{1}{\epsilon} \). Therefore, if we take \( N = \left\lceil \frac{1}{\epsilon} \right\rceil \), then for all \( n > N \), it follows that \( \left| \frac{\sin n}{n} \right| < \epsilon \).

Step 4: Conclude the Proof

Thus, the condition set by the definition of convergence is satisfied: for every \( \epsilon > 0 \), there exists a positive integer \( N \) (specifically, \( N = \left\lceil \frac{1}{\epsilon} \right\rceil \)) such that for all \( n > N \), \( \left| \frac{\sin n}{n} - 0 \right| < \epsilon \). Therefore, \( \lim_{n \to \infty} \frac{\sin n}{n} = 0 \).

Key Concepts

Convergence in Calculus

The concept of convergence in calculus is integral to understanding the behavior of sequences or functions as they tend toward a particular point. When we say a sequence converges, we mean it approaches a specific value, called the limit, as its terms progress indefinitely. This is foundational in calculus and analysis, serving as a bridge from discrete to continuous mathematics.

Convergence tells us that the sequence gets arbitrarily close to a limit, but not necessarily reach it. This idea is crucial when analyzing infinite sequences or series, as it describes how their sum or value behaves as the number of terms increases. In the given example, we're exploring how the sequence \( \frac{\sin n}{n} \) behaves as \( n \) grows without bound. The sequence converges to a limit because it approaches zero.

• It facilitates understanding the long-term behavior of functions and sequences.
• It provides the foundation for limits and continuity in calculus.
• It helps in determining convergence or divergence of series.

Understanding convergence is crucial for applying calculus to real-world problems, as it underpins many concepts such as integrals and derivatives.

Epsilon-Delta Definition

The epsilon-delta definition is a formal way to define the limit of a function or a sequence. It is one of the cornerstones of rigorous analysis in calculus. According to this definition, a sequence \( a_n \) converges to a limit \( L \) if for every positive number \( \epsilon \), there exists a positive integer \( N \) such that for all \( n > N \), the absolute difference \( |a_n - L| < \epsilon \) holds true.

Breaking it down into simpler terms:

• \( \epsilon \) represents how close we want to get to the limit \( L \).
• \( N \) is the threshold beyond which all terms of the sequence stay within \( \epsilon \) distance from \( L \).
• It ensures that as "n" becomes large, the function values get close to the limit and stay close.

This definition is essential because it forms a mathematically precise framework for understanding limits, allowing us to prove convergence definitively. Using the epsilon-delta definition, as seen in the step-by-step solution, ensures we can always verify whether a sequence truly converges to a limit. It's particularly useful in tackling sequences with complex behaviors, such as those involving trigonometric functions.

Trigonometric Limits

Trigonometric limits involve the behavior of trigonometric functions as their arguments approach a certain value or infinity. They are a specialized application of general limit rules and are crucial in many areas of calculus and trigonometric analysis.

Trigonometric functions such as \( \sin x \), \( \cos x \), and \( \tan x \) have unique properties that require special handling when approaching limits. For instance:

• Since the sine and cosine functions are bounded between -1 and 1, limits involving these functions often converge to particular values or zero.
• Combining trigonometric functions with algebraic expressions, as in \( \frac{\sin n}{n} \), often yields convergence to zero due to the algebraic term \( n \) growing much faster than the bounded trigonometric function.

In our example, the trigonometric limit \( \lim _{n \rightarrow \infty} \frac{\sin n}{n} = 0 \) showcases a classic behavior: as the denominator grows indefinitely, the ratio decreases towards zero. This understanding is vital in solving limits involving trigonometric functions, thus providing solutions to more intricate problems in calculus that require careful consideration of function behavior at specific values.