Problem 68 - University Calculus: Early Transcendentals

Question

Exploring \((\sin k x) / x \quad\) Graph \(y=(\sin x) / x, y=(\sin 2 x) / x,\) and \(y=(\sin 4 x) / x\) together over the interval \(-2 \leq x \leq 2 .\) Where does each graph appear to cross the \(y\) -axis? Do the graphs really intersect the axis? What would you expect the graphs of \(y=(\sin 5 x) / x\) and \(y=(\sin (-3 x)) / x\) to do as \(x \rightarrow 0 ?\) Why? What about the graph of \(y=(\sin k x) / x\) for other values of \(k ?\) Give reasons for your answers.

Step-by-Step Solution

Answer

Graphs approach the y-axis: \( y = \frac{\sin kx}{x} \) approaches \( k \) as \( x \to 0 \), following L'Hôpital's Rule.

Step 1: Understanding the Graphs

First, we need to understand how to graph the functions \( y = \frac{\sin x}{x}, \quad y = \frac{\sin 2x}{x}, \quad \text{and} \quad y = \frac{\sin 4x}{x} \).These functions are defined for all \( x eq 0 \) because division by zero is undefined.

Step 2: Graphing the Functions

We graph the functions over the interval \(-2 \leq x \leq 2\). Each of these functions resembles a damped sine wave, as the sine functions are divided by \(x\).- For \( y = \frac{\sin x}{x} \), the wave oscillates with a wavelength based on \(x\).- For \( y = \frac{\sin 2x}{x} \), the waves frequency doubles.- For \( y = \frac{\sin 4x}{x} \), the frequency quadruples.All graphs appear to behave similarly near the \(y\)-axis.

Step 3: Analyzing the Intersection with Y-axis

Each function \( \frac{\sin kx}{x} \) appears to have a removable discontinuity at \(x = 0\). From L'Hôpital's Rule or the limit, we find \( \lim_{x \to 0} \frac{\sin kx}{x} = k \), meaning each graph smoothly approaches the value \(k\) as \(x\) approaches 0 but doesn't touch the \(y\)-axis.

Step 4: Predicting Behavior of Other Functions

- For \( y = \frac{\sin 5x}{x} \), as \(x \to 0\), the limit is \(5\) due to the same reasoning.- For \( y = \frac{\sin (-3x)}{x} \), due to the odd property of sine function, its limit at \(x \to 0\) is \(-3\).Both expect to appear as touching but technically do not cross the \(y\)-axis.

Step 5: Generalizing for Other Values of k

For any function of the form \( \frac{\sin kx}{x} \), the limit as \(x \to 0\) is \(k\). This behavior holds for all \(k\), indicating graphs approach but don't intersect the \(y\)-axis at different heights, defined by \(k\). This aligns with the nature of \( \sin kx \).

Key Concepts

Sine Function

The sine function, denoted as \( \sin x \), is a periodic function with a standard period of \( 2\pi \). It ranges between -1 and 1. The sine of an angle in a right triangle represents the ratio of the length of the opposite side to the hypotenuse. When analyzing functions like \( \frac{\sin kx}{x} \), each graph alters the frequency of the standard sine wave. The parameter \( k \) affects the wave frequency directly:

• \( \sin x \) has a frequency of \( 1 \).
• \( \sin 2x \) doubles the frequency, creating more oscillations within the same interval.
• \( \sin 4x \) quadruples the frequency.

As \( x \) approaches zero, each function \( \frac{\sin kx}{x} \) forms a concentrated pattern around the y-axis. However, due to division by \( x \), they actually do not touch the y-axis but appear to converge closely.

L'Hôpital's Rule

When evaluating limits that result in an indeterminate form like \( \frac{0}{0} \), L'Hôpital's Rule is a powerful tool. This rule allows us to take the derivatives of the numerator and the denominator separately to potentially simplify the limit. Suppose you encounter \( \lim_{x \to 0} \frac{\sin kx}{x} \). Direct substitution fails as both the numerator \( \sin kx \) and denominator \( x \) tend towards zero.
Apply L'Hôpital's Rule by differentiating the sine function and \( x \):

• Derivative of \( \sin kx \) is \( k \cos kx \).
• Derivative of \( x \) is simply \( 1 \).

So, L'Hôpital's Rule gives us: \[ \lim_{x \to 0} \frac{\sin kx}{x} = \lim_{x \to 0} \frac{k \cos kx}{1} = k \cos 0 = k \]Hence, although the function appears undefined at zero, the limit provides the value that the function approaches, explaining the behavior near the y-axis.

Removable Discontinuity

A removable discontinuity occurs in a function when there is an apparent gap or hole in the graph for specific input values, yet limits exist around that point. It can often be 'removed' through algebraic simplification or by defining the function with that limit at the discontinuous point.
In the case of \( \frac{\sin kx}{x} \), the discontinuity at \( x = 0 \) is removable. The sine and cosine functions themselves are continuous across all real numbers. Therefore, the limit \( \lim_{x \to 0} \frac{\sin kx}{x} = k \) suggests the inclusion of a 'filled' value at zero, meaning it approaches, but never actually touches.

• This means graphically there’s no actual point at \( x = 0 \) unless defined by the limit.
• Introducing the limit fills this gap, making the discontinuity effectively vanish.

Thus, the notion of removable discontinuity highlights how the graphs behave as if intersecting the y-axis, while mathematically maintaining a well-defined limit close to it.