Problem 48 - Calculus

Question

For the functions in Problems \(46-53,\) do the following: (a) Make a table of values of \(f(x)\) for \(x=0.1,0.01,0.001\) \(0.0001,-0.1,-0.01,-0.001,\) and -0.0001 (b) Make a conjecture about the value of \(\lim _{x \rightarrow 0} f(x)\) (c) Graph the function to see if it is consistent with your answers to parts (a) and (b). (d) Find an interval for \(x\) near 0 such that the difference between your conjectured limit and the value of the function is less than \(0.01 .\) (In other words, find a window of height 0.02 such that the graph exits the sides of the window and not the top or bottom of the window.) $$f(x)=\sin 2 x$$

Step-by-Step Solution

Answer

The limit of \(f(x)\) as \(x\) approaches zero is 0.

Step 1: Create a Table of Values

Calculate the value of the function \( f(x) = \sin(2x) \) for the given values of \( x \). Get the sine of twice each input:- \( f(0.1) = \sin(0.2) = 0.1987 \)- \( f(0.01) = \sin(0.02) = 0.0200 \)- \( f(0.001) = \sin(0.002) = 0.0020 \)- \( f(0.0001) = \sin(0.0002) = 0.0002 \)- \( f(-0.1) = \sin(-0.2) = -0.1987 \)- \( f(-0.01) = \sin(-0.02) = -0.0200 \)- \( f(-0.001) = \sin(-0.002) = -0.0020 \)- \( f(-0.0001) = \sin(-0.0002) = -0.0002 \).These values suggest that as \( x \) approaches zero, \( f(x) \) also approaches zero.

Step 2: Conjecture the Limit

From the table of values, we observe that as \( x \) approaches zero, \( \sin(2x) \) approaches zero. Thus, we can conjecture that \( \lim_{x \to 0} \sin(2x) = 0 \).

Step 3: Graph the Function

Graph the function \( y = \sin(2x) \). Near \( x = 0 \), observe that the graph passes through the origin and approaches the x-axis on both sides, confirming our conjecture that the limit as \( x \) approaches zero is zero.

Step 4: Find the Interval Near Zero

We need an interval such that for any \( x \) in this interval, the function value differs from the limit zero by less than \( 0.01 \). The inequality \( \left| \sin(2x) - 0 \right| < 0.01 \) simplifies to \( \left| \sin(2x) \right| < 0.01 \). This requires \( \left| 2x \right| < 0.01 \), implying \( \left| x \right| < 0.005 \). Thus, the interval is approximately \((-0.005, 0.005)\).

Key Concepts

Trigonometric Functions

Understanding trigonometric functions is crucial when studying Calculus Limits, especially when dealing with function behavior as variables approach certain values. Trigonometric functions like sine and cosine are periodic and oscillate between a maximum and minimum.

In the exercise, we focus on the function \( f(x) = \sin(2x) \). The "2x" argument indicates that the sine wave will have a higher frequency compared to \( \sin(x) \), completing more cycles over the same interval. This means that as \( x \) approaches zero, \( \sin(2x) \) behaves closely like \( 2x \) due to the properties of sine near zero.

Key points about \( \sin(2x) \):

• Periodic function with a period of \( \pi \).
• Exhibits symmetric behavior around the y-axis.
• Behaves linearly near zero: \( \sin(2x) \approx 2x \).

This behavior is important in later steps of the solution, as it helps us approximate the limit of the function as \( x \) approaches zero.

Limit Conjecture

A limit conjecture is an educated guess about the value a function approaches as the input gets infinitely close to a certain point. In this exercise, we aim to determine \( \lim_{x \to 0} \sin(2x) \).

Based on the table of values calculated, you can see a clear pattern:

• As \( x \) becomes smaller (both positive and negative), \( \sin(2x) \) approaches zero.
• The values suggest that the closer you approach \( x = 0 \), the closer \( \sin(2x) \) gets to zero.

This consistent observation allows us to conjecture that the limit is zero.

When arriving at a conjecture, it's critical to employ analytical methods and validate the observations with graphical or numerical evidence. This conjecture, valid without divergence at \( x=0 \), is confirmed through subsequent analysis.

Graphical Analysis

Graphical analysis involves plotting the function to visually confirm the conjecture made regarding the limit. By graphing \( y = \sin(2x) \), we can visualize how the function behaves as \( x \) approaches zero.

When you look at this graph:

• The curve passes through the origin, consistent with \( \sin(0) = 0 \).
• Both to the left and right of \( x = 0 \), the graph approaches the horizontal axis, indicating the slope's direction (negative or positive) as \( x \) approaches zero.
• The oscillatory nature of sine doesn't interfere until you zoom in very close to zero.

This visual reaffirms that the limit is indeed zero. Graphs are powerful because they provide an intuitive understanding of the function's behavior, supporting our conjecture with ample visual evidence.

Function Intervals

The exercise requires finding an interval around zero where the function stays close to the conjectured limit. This helps in confirming numerical stability and correctness of the limit.

To determine such an interval, we need:

• The difference between \( \sin(2x) \) and zero must be less than \( 0.01 \).
• The condition \( \left| \sin(2x) \right| < 0.01 \) is assessed to find the suitable interval.

We simplify this to \( \left| 2x \right| < 0.01 \), resulting in the interval \( (-0.005, 0.005) \). This interval means:

• For any \( x \) within this range, \( \sin(2x) \) will deviate by less than \( 0.01 \) from zero.
• The graph of \( \sin(2x) \) within this interval lies between two horizontal lines at \( y = 0.01 \) and \( y = -0.01 \), maintaining proximity to the conjectured limit.

This approach ensures both mathematical precision and visual accuracy.