Problem 15 - 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. $$ f(x)=x+1, \quad L=5, \quad c=4, \quad \epsilon=0.01 $$

Step-by-Step Solution

Answer

The open interval is \(3.99 < x < 4.01\) and \(\delta = 0.01\).

Step 1: Understand the Given Function

The problem provides the function \( f(x) = x + 1 \). We know that at \( x = c \) where \( c = 4 \), the function \( f(x) \) is close to \( L = 5 \). We need to verify that this holds for a certain interval of \( x \) around \( c \).

Step 2: Evaluate the Function at c

First, let's compute \( f(c) \). For \( c = 4 \), we get \( f(4) = 4 + 1 = 5 \). This confirms that \( f(c) = L \).

Step 3: Setup the Inequality

The goal is to find an interval around \( c = 4 \) such that \(|f(x) - L| < \epsilon \). We substitute \( f(x) = x + 1 \), \( L = 5 \), and \( \epsilon = 0.01 \) into the inequality: \ \(|(x + 1) - 5| < 0.01\).

Step 4: Simplify the Inequality

Simplify \(|x + 1 - 5| < 0.01\) to \(|x - 4| < 0.01\). This tells us that the interval around \( c = 4 \) is such that \( x \) must satisfy this inequality.

Step 5: Determine the Open Interval

From the inequality \(|x - 4| < 0.01\), we derive the interval: \ \( 3.99 < x < 4.01 \). This is the open interval where the condition \(|f(x) - L| < \epsilon\) holds.

Step 6: Establish a Value for δ

Since \(|x - 4| < 0.01\) already gives us the desired inequality for \(|f(x) - L| < \epsilon\), we set \( \delta = 0.01 \). Thus, for any \( x \) such that \( 0 < |x - 4| < 0.01 \), it follows that \(|f(x) - 5| < 0.01\).

Key Concepts

Limit of a Function

A limit in mathematics describes the value that a function approaches as the input (or variable) approaches some value. To understand limits, it's crucial to focus on what value the function heads towards rather than what the function equals at that point. In our exercise, when talking about the limit, we're saying that the value of the function \(f(x)\) should get as close as possible to \(L = 5\) as \(x\) gets closer to \(c = 4\).
This idea forms the basis of calculus and helps us deal with values that approach specific numbers. It's like watching a car get closer to a parking spot without actually reaching it. The fact that \(f(4) = 5\) simply reinforces that the function behaves well at \(x = 4\), but our main concern is ensuring this behavior holds in the neighborhood around 4.
Understanding limits helps us predict the behavior of functions near certain points, which is essential in calculus for determining continuity and evaluating derivatives.

Open Interval

An open interval is a range of numbers that does not include its endpoints. This concept is especially useful when we discuss limits, as it helps define the range where a condition holds true. In our exercise, the open interval describes the values of \(x\) that keep \(|f(x)-L|\) below \(\epsilon\).

• For the function \(f(x) = x + 1\) with \(L = 5\) and \(\epsilon = 0.01\), the open interval is derived from the inequality \(|x - 4| < 0.01\).
• This inequality means that \(x\) must be strictly greater than 3.99 and less than 4.01, or \(3.99 < x < 4.01\).
  

Open intervals are fundamental when dealing with limits because they provide the precise range of continuity and ensure that we get arbitrarily close to the limit without touching the boundary values. They give us "breathing room" to explore how functions behave as they gently tiptoe around specific points.

Inequality Solution

Solving inequalities is crucial when dealing with limits because it helps us establish the range of values where the function behaves as desired. In this exercise, we're working with the inequality \(|x - 4| < 0.01\), and solving it is straightforward but vital.

• The inequality \(|x - 4| < 0.01\) translates to two simple conditions: \(x - 4 < 0.01\) and \(x - 4 > -0.01\).
• Solving these gives us \(x < 4.01\) and \(x > 3.99\), thus the solution \(3.99 < x < 4.01\).

This solution helps identify the exact range where the function \(f(x)\) stays within \(\epsilon\) units from \(L\). It's a neat combination of understanding and applying inequalities that guarantees precise control over the behavior of the function. Using inequalities like this reinforces the precision and power of calculus, ensuring that we can handle even the smallest closeness criteria defined by \(\epsilon\).