Problem 65 - Precalculus: Functions and Graphs

Question

Graph the equation on the Interval \([-2,2]\), and describe the behavior of \(y\) as \(x \rightarrow 0^{-}\) and as \(x \rightarrow 0^{+}\) \(y=\frac{\sin 2 x}{x}\)

Step-by-Step Solution

Answer

As \( x \to 0^{-} \) and \( x \to 0^{+} \), \( y = \frac{\sin 2x}{x} \) approaches 2.

Step 1: Identify the Function and Interval

We need to graph the function \( y = \frac{\sin 2x}{x} \) in the interval \([-2, 2]\). This interval tells us the range of \( x \) values over which we need to visualize the graph.

Step 2: Analyze the Continuity at x=0

The function \( y = \frac{\sin 2x}{x} \) is not defined at \( x = 0 \) because it results in a division by zero. To analyze the behavior near \( x = 0 \), we consider the limit as \( x \) approaches zero.

Step 3: Calculate the Limit as x Approaches 0

Using L'Hôpital's Rule (since \( \frac{\sin 2x}{x} \) is an indeterminate form \( \frac{0}{0} \) as \( x \rightarrow 0 \)), we find the limit: \[ \lim_{x \to 0} \frac{\sin 2x}{x} = \lim_{x \to 0} \frac{2\cos 2x}{1} = 2 \cos(0) = 2. \]Thus, the behavior of the function as \( x \to 0 \) from either side tends to 2.

Step 4: Sketch the Graph

Start by plotting key points in the interval \([-2, 2]\), such as \( x = -2, \frac{-3}{2}, -1, 0, 1, \frac{3}{2}, 2 \). Calculate \( y \) (except at 0) for each. For the point where \( x = 0 \), use the limit we calculated to know the function approaches \( y = 2 \). Sketch how the curve smoothly moves through these points, keeping in mind the asymptotic approach to \( y = 2 \) near \( x = 0 \). Note that there is a removable discontinuity at \( x = 0 \).

Step 5: Describe Behavior as x Approaches 0 from Both Sides

As \( x \rightarrow 0^{-} \), the function \( y = \frac{\sin 2x}{x} \) approaches 2 from the left of zero due to our limit result. Similarly, as \( x \rightarrow 0^{+} \), \( y \) approaches 2 from the right. This implies \( y \) is continuous around zero after removing the discontinuity with the limit value.

Key Concepts

Removable Discontinuity

A removable discontinuity occurs in a function when a certain point is undefined or doesn't fit smoothly into the function, but it can be "removed" by redefining the function at that point. Essentially, it's like a hole in the graph that can be filled by adjusting the function's definition. In our problem, the function

• \( y = \frac{\sin 2x}{x} \)
• is not defined at \( x = 0 \) since this leads to division by zero.
• However, using limits, we can determine the value that the function "wants" to be at \( x = 0 \), which is 2.

By introducing this limit value into the graph as a point at \( x = 0 \), we effectively remove the discontinuity. This helps create a smoother and more complete graph by filling in that point.

L'Hôpital's Rule

L'Hôpital's Rule is a handy tool for finding limits of indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). In our exercise, the function \( y = \frac{\sin 2x}{x} \) presented such an indeterminate form at \( x = 0 \). Here's how L'Hôpital's Rule helps:

• We determine that both the numerator \(\sin 2x\) and the denominator \(x\) approach 0 as \(x \rightarrow 0\).
• We apply L'Hôpital's Rule by differentiating the numerator and denominator:
• The derivative of \(\sin 2x\) is \(2\cos 2x\), and the derivative of \(x\) is \(1\).

The limit using these derivatives becomes \[\lim_{x \to 0} \frac{2\cos 2x}{1} = 2 \cos(0) = 2.\]This demonstrates that as \(x\) approaches 0, the function approaches a value of 2, helping us remove the discontinuity.

Limit of a Function

The concept of a limit helps us understand the behavior of functions as they approach specific points. When dealing with limits, especially where a function isn't defined at a point, the idea is to see what value the function is trending toward as it gets infinitely close. In our situation:

• We look at \( y = \frac{\sin 2x}{x} \) as \( x \) approaches 0 from both sides.
• The previous calculation using L'Hôpital's Rule shows that the limit is 2, even though the function is not directly defined at \( x = 0 \).
• This trend toward 2 indicates the point of the removable discontinuity if incorporated into the graph, creating a seamless path around \( x = 0 \).

Thus, limits are crucial for analyzing the behavior of functions around problem spots where they aren't defined.

Continuity of Functions

Continuity is all about a function being smooth without breaks, holes, or jumps. A function is continuous at a point if it is defined at that point, and its limit exists and corresponds to the function's value. In the example we worked through:

• The function \( y = \frac{\sin 2x}{x} \) is initially not defined at \( x = 0 \), creating a continuity concern.
• However, we found through calculating the limit that the function behaves as though its value is 2 when close to \(x = 0\) from either the left or right.
• By considering this limit value as the function's value at \( x = 0 \), the discontinuity is removed, restoring continuity.

Thus, the concept of continuity is tied closely to limits and removable discontinuities, ensuring the graph flows seamlessly without interruptions.