Problem 20 - Thomas Calculus

Question

Each of Exercises \(15-30\) gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon>0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. $$ f(x)=\sqrt{x-7}, \quad L=4, \quad c=23, \quad \epsilon=1 $$

Step-by-Step Solution

Answer

The open interval is (16, 32) and \( \delta = 7 \).

Step 1: Understand the Function and Parameters

The function given is \( f(x) = \sqrt{x-7} \). We need to ensure the inequality \(|f(x) - L| < \epsilon \) is satisfied, where \( L = 4 \) and \( \epsilon = 1 \). This translates to \( |\sqrt{x-7} - 4| < 1 \).

Step 2: Solve the Inequality

Starting with the inequality \( |\sqrt{x-7} - 4| < 1 \), we can express it as two separate inequalities: 1. \( \sqrt{x-7} - 4 < 1 \) 2. \( \sqrt{x-7} - 4 > -1 \). Solving the first inequality: \( \sqrt{x-7} < 5 \) implies \( x-7 < 25 \), so \( x < 32 \). Solving the second inequality: \( \sqrt{x-7} > 3 \) implies \( x-7 > 9 \), so \( x > 16 \). The solution to \( |\sqrt{x-7} - 4| < 1 \) is thus \( 16 < x < 32 \).

Step 3: Find \(\delta\) for \(0<|x-c|<\delta\)

Given \( c = 23 \), we are looking for values for \( \delta \) such that \( 0 < |x-23| < \delta \) leads to \( 16 < x < 32 \). We find the extra boundary conditions based on \( c=23 \) by figuring out how far 23 is from the extremes of the interval 16 and 32. \( |23 - 16| = 7 \) and \( |23 - 32| = 9 \). The valid \( \delta \) is the smaller of these distances, so \( \delta = 7 \).

Step 4: Conclude with Open Interval and \(\delta\)

The open interval about \( c = 23 \) where the inequality \(|f(x) - L| < \epsilon\) holds is \( (16, 32) \). Thus, for all \( x \) such that \( 0 < |x - 23| < \delta \), \( |\sqrt{x-7} - 4| < 1 \). The value of \( \delta \) for which this is true is \( \delta = 7 \).

Key Concepts

Epsilon-Delta Definition

In calculus, the epsilon-delta definition of limits is fundamental to understanding limits and continuity. It helps us rigorously prove that a function approaches a certain value, known as the limit, as the input approaches a specific point. In this exercise, we apply the epsilon-delta definition to the function \( f(x) = \sqrt{x-7} \) with an expected limit \( L = 4 \) at \( c = 23 \).
This involves finding a \( \delta > 0 \) for any given \( \epsilon > 0 \). Meaning, for every \( \epsilon \), no matter how small, there's a corresponding \( \delta \) such that if the distance between a point \( x \) and \( c \) is less than \( \delta \), then the distance between \( f(x) \) and \( L \) is less than \( \epsilon \).

• \( \epsilon \) controls how close \( f(x) \) needs to be to the limit \( L \).
• \( \delta \) determines how close \( x \) must be to \( c \) to keep \( f(x) \) within \( \epsilon \) units of \( L \).

Here, \( \epsilon = 1 \), so we found an interval where \(|\sqrt{x-7} - 4| < 1\). Solving this helped determine delta, ensuring \(|f(x) - L|< \epsilon\) is satisfied within an open interval.

Continuity of Functions

Continuity is a property of functions that shows that the function does not have any 'jumps' or 'breaks'. A function \( f(x) \) is continuous at a point \( x = c \) if the following condition holds:

• The limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \).

For the function \( f(x) = \sqrt{x-7} \), we tested for continuity at \( c = 23 \). This requires that as \( x \) gets closer and closer to 23, \( f(x) \) approaches its value at 23 smoothly without interruption.
Ensuring continuity involves confirming the interval \( x \in (16, 32) \), as derived from solving the inequality. Within this interval, the function is defined and behaves predictably. Therefore, the concept of continuity and the epsilon-delta condition are nicely aligned to verify that \( f(x) \) doesn't have sudden changes near \( c = 23 \).

Inequality Solutions

Solving inequalities is crucial to finding the intervals in which certain properties of functions hold true. In this exercise, we solved the inequality \(|\sqrt{x-7} - 4| < 1\), starting by breaking it down into two separate inequalities:

• \( \sqrt{x-7} - 4 < 1 \)
• \( \sqrt{x-7} - 4 > -1 \)

By solving these, we found the range of \( x \) values that ensure the function's output stays within \( \epsilon = 1 \) of \( L = 4 \). For \( \sqrt{x-7} - 4 < 1 \), it simplifies to \( x < 32 \), and for \( \sqrt{x-7} - 4 > -1 \), it becomes \( x > 16 \).
Thus, the solution comes from the intersection of these intervals, ultimately giving \( 16 < x < 32 \). This is the open interval about \( c = 23 \) that ensures the function's closeness to the limit is within the acceptable range defined by \( \epsilon \). Solving these inequalities is pivotal to applying the epsilon-delta definition effectively.