Problem 17 - 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. $$ f(x)=\sqrt{x+1}, \quad L=1, \quad c=0, \quad \epsilon=0.1 $$

Step-by-Step Solution

Answer

The open interval is \((-0.19, 0.21)\) and \(\delta = 0.19\).

Step 1: Calculate the inequality

To find the open interval for \[ |f(x) - L| < \epsilon, \]we start with the function given \[ f(x) = \sqrt{x+1}. \]Plugging this into the inequality, we have:\[ |\sqrt{x+1} - 1| < 0.1. \]

Step 2: Set up a new inequality

Simplifying \[ |\sqrt{x + 1} - 1| < 0.1 \] leads us to \[ -0.1 < \sqrt{x + 1} - 1 < 0.1 \] and thus,\[ 0.9 < \sqrt{x + 1} < 1.1. \]

Step 3: Remove the square root

To simplify further, square all parts of the inequality:\[ (0.9)^2 < x + 1 < (1.1)^2. \] Calculating gives:\[ 0.81 < x + 1 < 1.21. \]

Step 4: Solve for x

Subtract 1 from each part of the inequality to isolate x:\[ 0.81 - 1 < x < 1.21 - 1, \] which simplifies to\[ -0.19 < x < 0.21. \]

Step 5: Choose a value for δ

This tells us that the open interval where \[ |f(x)-L| < \epsilon = 0.1 \] is \[ (-0.19, 0.21). \] To ensure \[ 0 < |x - c| < \delta \] implies \[ |f(x) - L| < \epsilon, \] choose \[ \delta = 0.19. \]

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a formal way to describe the limit of a function. It ensures that the function gets as close as desired to a particular value as the input approaches a certain point. To understand this, let’s focus on the components:

• \( \epsilon \) (epsilon) represents how close we want the function value \( f(x) \) to be to the limit \( L \).
• \( \delta \) (delta) represents how close \( x \) needs to be to \( c \) (the point of interest) to achieve this closeness.

The goal is to find a \( \delta \) such that whenever \( 0 < |x - c| < \delta \), it follows that \( |f(x) - L| < \epsilon \). This means we are creating a control on \( x \) so that \( f(x) \) remains within an \( \epsilon \) distance from \( L \).
In our exercise, \( L = 1 \), \( c = 0 \), and \( \epsilon = 0.1 \). By determining a suitable \( \delta \), we ensure that \( f(x) = \sqrt{x+1} \) remains less than 0.1 away from 1 as \( x \) gets close to 0.

Open Interval

An open interval is a range of numbers between two endpoints where the endpoints themselves are not included. In mathematics, an open interval is usually denoted as \( (a, b) \), meaning all numbers between \( a \) and \( b \) but not \( a \) or \( b \) themselves.In the context of this exercise, we are asked to determine an open interval around \( c = 0 \) where the inequality \( |f(x) - L| < \epsilon \) holds. By solving the inequality, we find:

• The range where the condition is satisfied, resulting in \( -0.19 < x < 0.21 \).

This defines our open interval as \( (-0.19, 0.21) \). This means that any \( x \) value within this range will ensure the function \( \sqrt{x+1} \) stays within \( 0.1 \) units from \( 1 \). The endpoints of the interval \( -0.19 \) and \( 0.21 \) are not part of the interval. Therefore, \( f(x) \) when \( x = -0.19 \) or \( 0.21 \) would actually reach the boundaries of the given \( \epsilon \).

Inequality Solving

Inequality solving is about finding the range of values that satisfy a certain condition. In this exercise, we need to manipulate the expression \( |\sqrt{x+1} - 1| < 0.1 \).Here's a concise method to solve it:

• First, express the absolute inequality |\( |\sqrt{x+1} - 1| < 0.1 \) as a double inequality: \( -0.1 < \sqrt{x+1} - 1 < 0.1 \).
• Then, isolate the square root by adding 1: \( 0.9 < \sqrt{x+1} < 1.1 \).
• To remove the square root, square all parts: \( (0.9)^2 < x + 1 < (1.1)^2 \).
• Calculate to find: \( 0.81 < x + 1 < 1.21 \).
• Finally, isolate \( x \) by subtracting 1 from everything: \( -0.19 < x < 0.21 \).

This gives us the interval where the inequality holds, ensuring \( f(x) \) is close to \( L \) as required by the \( \epsilon \) condition. Solving inequalities is crucial for determining the valid range of \( x \) within certain mathematical conditions. It helps pinpoint where a function behaves predictably.