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Chapter 5: Problem 10

Prove that the limit \(\lim _{x \rightarrow 0} \sin (1 / x)\) fails to exist.

Short Answer

Expert verified
The limit does not exist because \( \sin(1/x) \) oscillates indefinitely as \( x \rightarrow 0 \).

Step by step solution

01

Understanding the Limit Problem

We are asked to determine whether the limit \( \lim_{x \rightarrow 0} \sin(1/x) \) exists. This involves analyzing the behavior of \( \sin(1/x) \) as \( x \) approaches 0 from both positive and negative directions.
02

Identifying the Function's Behavior

As \( x \) approaches 0, \( 1/x \) becomes very large, making \( \sin(1/x) \) oscillate increasingly rapidly between -1 and 1. This rapid oscillation suggests instability in reaching a single value.
03

Contradiction through Subsequence

To prove the limit doesn't exist, we can show that there are subsequences approaching 0 that result in different limit values. Consider \( x_n = rac{1}{ rac{\pi}{2} + 2n\pi} \) and \( y_n = rac{1}{2n\pi} \). For \( x_n \), \( \sin(1/x_n) = 1 \) and for \( y_n \), \( \sin(1/y_n) = 0 \). Both subsequences approach 0, yet produce different limit behaviors.
04

Conclusion: Non-Existence of the Limit

Since different subsequences \( x_n \) and \( y_n \) of \( x \) approaching zero lead to different limits for \( \sin(1/x) \), the original limit \( \lim_{x \rightarrow 0} \sin(1/x) \) fails to exist.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Behavior of sin(1/x)
The function \( \sin(1/x) \) can be quite intriguing as \( x \) approaches zero. As \( x \) gets closer to 0, the value of \( 1/x \) becomes very large. This, in turn, impacts the sine function, which takes the output of \( 1/x \) and maps it to values between -1 and 1. In this scenario, we see
  • As \( x \to 0^+ \), \( \sin(1/x) \) oscillates rapidly, creating a complex pattern.
  • As \( x \to 0^- \), similar rapid oscillations occur, causing the function to swing between -1 and 1 continuously.
This rapid oscillation indicates that \( \sin(1/x) \) does not stabilize at a single value, making it difficult to predict what the function approaches as \( x \) nears zero.
Considering these oscillations suggests that capturing a single definitive limit can be challenging.
The Role of Oscillation in Limits
Oscillation refers to a function's repeated rise and fall. \( \sin(1/x) \) exemplifies this as we observe:
  • The sine wave naturally oscillates between -1 and 1. This is inherent in any sine function regardless of its input.
  • When the input to the sine function, like \( 1/x \), becomes large, it results in more cycles of oscillation within a small interval of \( x \) values.
  • As \( x \to 0 \), these oscillations become quicker and more condensed, reinforcing instability.
Such behavior indicates that at \( x = 0 \), there is no tendency for \( \sin(1/x) \) to settle at a single value. Thus, the function continuously oscillates without converging to a precise limit.
This continuous overload of oscillations signals the non-existence of a limit when \( x \to 0 \).
Exploring Non-existence of Limits
The concept of a limit fundamentally relies on the idea that a function will settle at a particular value as its input approaches a certain point. For \( \sin(1/x) \), given its oscillatory nature:
  • We can observe different subsequences approaching zero that lead to different results.
  • For instance, take sequences like \( x_n = \frac{1}{\frac{\pi}{2} + 2n\pi} \) which make \( \sin(1/x_n) \to 1 \).
  • Another sequence \( y_n = \frac{1}{2n\pi} \) results in \( \sin(1/y_n) \to 0 \).
Both subsequences converge to 0, yet produce distinct result trends for \( \sin(1/x) \). This serves as a key demonstration that a uniform limit does not exist. Simply stated, if subsequences yield different limits, the primary sequence cannot have a definite limit.
Ultimately, this highlights that \( \lim_{x \rightarrow 0} \sin(1/x) \) does not exist due to the lack of convergence at a single value.

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Most popular questions from this chapter

Prove that the function \(f(x)=x^{2}\) is continuous at every point of \(\mathbb{R}\) using the \(\delta-\varepsilon\) form of continuity,

Prove the validity of the limit $$ \lim _{x \rightarrow x_{0}}(a x+b)=a x_{0}+b $$ by converting to a statement about sequences.

A function \(f: E \rightarrow \mathbb{R}\) is said to be Lipschitz if there is a positive number \(M\) so that \(|f(x)-f(y)| \leq M|x-y|\) for all \(x, y \in E\). Show that such a function must be uniformly continuous on \(E .\) Is the converse true?

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a continuous function that is periodic in the sense that for some number \(p, f(x+p)=f(x)\) for all \(x \in \mathbb{R}\). Show that \(f\) has an absolute maximum and an absolute minimum.

Give three examples of a function \(f\) that fails to be continuous at a point \(x_{0}\). The first should be discontinuous merely because \(f\) is not defined at \(x_{0}\). The second should be discontinuous because \(\lim _{x \rightarrow x_{0}} f(x)\) fails to exist. The third should have neither of these defects but should nonetheless be discontinuous.
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