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Magazine Chapter 5: Problem 65
Graph the equation on the Interval \([-2,2]\), and describe the behavior of \(y\) as \(x \rightarrow 0^{-}\) and as \(x \rightarrow 0^{+}\) \(y=\frac{\sin 2 x}{x}\)
Short Answer
Expert verified
As \( x \to 0^{-} \) and \( x \to 0^{+} \), \( y = \frac{\sin 2x}{x} \) approaches 2.
Step by step solution
01
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.
02
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.
03
Calculate the Limit as x Approaches 0
Using L'H么pital'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.
04
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 \( x = 0 \).
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
L'H么pital's Rule
L'H么pital'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'H么pital's Rule helps:
- We determine that both the numerator \(\sin 2x\) and the denominator \(x\) approach 0 as \(x \rightarrow 0\).
- We apply L'H么pital's Rule by differentiating the numerator and denominator:
- The derivative of \(\sin 2x\) is \(2\cos 2x\), and the derivative of \(x\) is \(1\).
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'H么pital'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 \( x = 0 \).
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.
