Problem 75 - Calculus

Question

Consider the function \(f(x)=\sin (1 / x)\) (a) Find a sequence of \(x\) -values that approach 0 such that \(\sin (1 / x)=0\) [Hint: Use the fact that \(\sin (\pi)=\sin (2 \pi)=\) \(\sin (3 \pi)=\ldots=\sin (n \pi)=0 .]\) (b) Find a sequence of \(x\) -values that approach 0 such that \(\sin (1 / x)=1\) IHint: Use the fact that \(\sin (n \pi / 2)=1\) if \(n=\) \(1,5,9, \dots .]\) (c) Find a sequence of \(x\) -values that approach 0 such that \(\sin (1 / x)=-1\) (d) Explain why your answers to any two of parts (a)-(c) show that \(\lim _{x \rightarrow 0} \sin (1 / x)\) does not exist.

Step-by-Step Solution

Answer

The limit does not exist because sequences approaching zero yield different sine values (0, 1, -1).

Step 1: Sequence for zero sine value

Consider the sequence \[ x_n = \frac{1}{n\pi}, \text{where } n \text{ is a positive integer}. \]For each value of \(x_n\), we have \[ 1/x_n = n\pi. \]Since \(\sin(n\pi) = 0\), this sequence will satisfy \(\sin(1/x_n)=0\). As \(n\) approaches infinity, \(x_n\) approaches zero.

Step 2: Sequence for sine value equal to one

Consider the sequence \[ x_m = \frac{2}{(4m-1)\pi}, \text{where } m \text{ is a positive integer}. \]For each value of \(x_m\), we have \[ 1/x_m = (4m-1)\frac{\pi}{2}. \]Since \(\sin((4m-1)\frac{\pi}{2}) = 1\), this sequence will satisfy \(\sin(1/x_m) = 1\). As \(m\) approaches infinity, \(x_m\) approaches zero.

Step 3: Sequence for sine value equal to negative one

Consider the sequence \[ x_k = \frac{2}{(4k+1)\pi}, \text{where } k \text{ is a positive integer}. \]For each value of \(x_k\), we have \[ 1/x_k = (4k+1)\frac{\pi}{2}. \]Since \(\sin((4k+1)\frac{\pi}{2}) = -1\), this sequence will satisfy \(\sin(1/x_k) = -1\). As \(k\) approaches infinity, \(x_k\) approaches zero.

Step 4: Explain why the limit does not exist

To prove the non-existence of the limit\[ \lim_{x \to 0} \sin(1/x) \] consider three sequences: from step 1, \(\sin(1/x_n) = 0\); from step 2, \(\sin(1/x_m) = 1\); and from step 3, \(\sin(1/x_k) = -1\). All sequences approach \(x = 0\), but they lead to different function values (0, 1, and -1 respectively). Therefore, the limit does not exist.

Key Concepts

Limit of a function

The limit of a function examines the behavior of the function as the input values get closer to a particular value. Here, we consider the limit of the function \(f(x) = \sin(1/x)\) as \(x\) approaches zero. A limit exists if the function approaches a specific value as \(x\) approaches the target value from any direction. However, if the function does not settle to one value, then the limit does not exist.
In this exercise, several sequences approach zero, but \(\sin(1/x)\) yields different results for each sequence. This discrepancy indicates that \(\lim_{x \to 0} \sin(1/x)\) does not exist. Thus, it's essential to recognize the behavior of functions around critical points to understand the concept of limits clearly.

Sequences

Sequences are ordered lists of numbers that often converge to a point as they progress. In calculus, sequences play a crucial role because they can be used to evaluate the limit of a function.
In this example, we analyzed sequences for \(x\) values that approach 0, ensuring that different trigonometric values resulted from \(\sin(1/x)\). These sequences were:

• \(x_n = \frac{1}{n\pi}\): where \(n\) is an integer, leading the sequence \(\sin(1/x_n) = 0\).
• \(x_m = \frac{2}{(4m-1)\pi}\): leading to the sequence \(\sin(1/x_m) = 1\).
• \(x_k = \frac{2}{(4k+1)\pi}\): resulting in the sequence \(\sin(1/x_k) = -1\).

Each of these sequences approaches zero but results in different sine values, thereby demonstrating the function's complex behavior around zero and the implications for the non-existence of the limit.

Trigonometric functions

Trigonometric functions like sine, cosine, and tangent are fundamental in calculus and describe relationships in triangles. The sine function specifically relates to angles and their respective lengths in a right-angled triangle, but it also exhibits periodic behavior, meaning it repeats its values every \(2\pi\) units.
The exercise taps into this periodic nature of the sine function, noting special angles where specific sine values occur:

• \(\sin(n\pi) = 0\) for whole number \(n\).
• \(\sin(\frac{(4m-1)\pi}{2}) = 1\).
• \(\sin(\frac{(4k+1)\pi}{2}) = -1\).

Understanding these characteristics allows us to create sequences approaching different trigonometric outputs and highlights the non-continuous nature of \(\sin(1/x)\) at \(x\to0\). This insight into trigonometric functions is essential to tackling complex calculus problems.

Non-existent limit

A non-existent limit occurs when a function does not converge to a single value as the input approaches a specific number. In this exercise, \(\sin(1/x)\) demonstrates this phenomenon as it instead approaches multiple outcomes (0, 1, and -1) depending on different paths taken towards zero.
This non-uniform approach is due to the oscillating nature of \(\sin(1/x)\), which does not settle down to a single number. Each path (or sequence) gives a valid outcome but diverges from the others, proving that without a consistent value across all paths, the limit \(\lim_{x \to 0} \sin(1/x)\) cannot exist.
By delving into such problems, students gain a deeper understanding of how limits work, and the reasons behind the existence or non-existence of a limit, fostering a robust intuition behind calculus concepts.