Question

In the following problems, find the limit of the given sequence as \(n \rightarrow \infty\). $$n \sin (1 / n)$$

Short answer

The limit is 1.

Step-by-step solution

Understand the Problem

We need to find the limit of the sequence \rightarrow \rightarrow \rightarrow \rightarrow \to \to \rightarrow \rightarrow \rightarrow } as \(n\) approaches \(\rightarrow n \rightarrow \to \rightarrow \). This involves analyzing the behavior of the expression as \(n\) becomes very large.

Rewrite the Expression

Rewrite \(n \sin(\frac{1}{n})\) in a more manageable form. Notice that when \(n\) is large, \(\frac{1}{n}\) becomes very small. Thus, the expression inside the sine function, \(\sin(\frac{1}{n})\), can be approximated by its Taylor series expansion.

Use the Taylor Series Expansion

For small values of \(x\), the Taylor series expansion of \(\sin(x)\) around 0 is approximately \(\sin(x) \approx x\). Therefore, we can approximate \(\sin \left( \frac{1}{n} \right)\) as \(\frac{1}{n}\).

Simplify the Expression

Substitute the approximation into the sequence: \(n \sin \left( \frac{1}{n} \right) \approx n \left( \frac{1}{n} \right) = 1\).

Find the Limit

Now that the sequence simplifies to a constant: \(n \sin \left( \frac{1}{n} \right) \approx 1\). The limit of a constant as \(n\) approaches infinity is just the constant itself. Therefore, the limit is 1.

Key concepts

Taylor series expansion

Taylor series expansion is a powerful tool in mathematics to approximate functions. It expresses a function as an infinite sum of terms calculated from its derivatives at a single point. This series gives us a polynomial approximation of functions which is very useful for simplifying complex expressions.
In the context of our problem, where we need to find the limit of the sequence as \(n \rightarrow \infty\), we approximate the sin function using its Taylor series.

• For small values of \(x\), the Taylor series of \(\sin(x)\) around zero is:

\[ \sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \quad \text{(only keeping the first term for small x)} \]
When \(x\) is very small, higher-order terms like \(x^3\) become negligible, and \(\sin(x)\) approximates to \(x\). This simplification helps break down complex functions into workable linear forms.
Using Taylor series in this manner allows us to approximate \(\sin(1/n)\) as \(1/n\) because \(1/n\) is very small for large \(n\).

Infinity

Infinity is a key concept in calculus and analysis. It represents values that grow larger and larger without bound. When working with sequences and limits, we often explore the behavior of functions as they approach infinity.
In our sequence, we are interested in what happens as \(n\) grows infinitely large. Understanding this concept is crucial because it helps us deduce that small quantities like \(1/n\) approach zero as \(n\) approaches infinity.
As \(n\) becomes larger and larger:

• \(1/n\) becomes smaller and smaller, approaching 0.

Thus, infinity allows us to analyze the limit of sequences and expressions more easily.
By examining how the sequence \(n\sin(1/n)\) behaves as \(n\) grows extremely large, we determine that the effect of \(\sin(1/n)\) approximates to \(1/n\), and therefore our initial expression tends to a constant value.
If we rewrite the expression with this approximation:

• \(n \sin(1/n) \approx n \cdot (1/n) = 1\)

The limit as \(n \rightarrow \infty\) for our sequence is clearly seen to be 1.

Sin function approximation

The sine function, \(\sin(x)\), is foundational in trigonometry and calculus. When we approximate \(\sin(x)\) for very tiny values of \(x\), we leverage its Taylor series expansion, which simplifies the analysis of trigonometric functions.
From the initial problem, \(n \sin(1/n)\), we note that as \(n\) becomes very large, \(1/n\) becomes very small. Using our approximation:

• \(\sin(1/n) \approx 1/n\)

This simplification is detailed by the Taylor series where:\[ \sin(x) \approx x \quad \text{for very small}\ x \]
As a result, substituting this approximation back into the sequence:

• \(n \sin(1/n) \approx n \cdot 1/n = 1\)

This method shows that the limit of the sequence \(n \sin(1/n)\) as \(n \rightarrow \infty\) is 1. Understanding sin function approximation lets us simplify evaluations and solve limits more easily.