Problem 22 - 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. $$ f(x)=x^{2}, \quad L=3, \quad c=\sqrt{3}, \quad \epsilon=0.1 $$

Step-by-Step Solution

Answer

The interval is approximately \((1.7029, 1.7607)\) and \(\delta = 0.0287\).

Step 1: Understand the Problem

We need to find an open interval around \(c = \sqrt{3}\) where the function \(f(x) = x^2\) is within \(\epsilon = 0.1\) of \(L = 3\). This means we need \(|f(x) - L| < \epsilon\), which simplifies to \(|x^2 - 3| < 0.1\).

Step 2: Set Up the Inequality

The inequality \(|x^2 - 3| < 0.1\) can be rewritten as \(-0.1 < x^2 - 3 < 0.1\). This further simplifies to \(2.9 < x^2 < 3.1\).

Step 3: Solve the Inequality for x

We solve \(2.9 < x^2 < 3.1\) by taking square roots. The square roots give us \(\sqrt{2.9} < x < \sqrt{3.1}\), approximately \(1.7029 < x < 1.7607\). So, the interval is \((1.7029, 1.7607)\).

Step 4: Translate to a \(\delta\) Expression

Now, we need \(0 < |x - \sqrt{3}| < \delta\). Since \(c = \sqrt{3}\), substitute \(\approx 1.732\), so \(\delta\) is the minimum distance from \(1.732\) to the bounds of \((1.7029, 1.7607)\). Thus, \(\delta\) is the smaller of \(1.732 - 1.7029\) and \(1.7607 - 1.732\).

Step 5: Calculate \(\delta\)

Calculate the distances: \(1.732 - 1.7029 = 0.0291\) and \(1.7607 - 1.732 = 0.0287\). The smaller value is \(0.0287\), hence \(\delta = 0.0287\).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a formal way to describe the limit of a function. It is a cornerstone of calculus and is vital in proving that a limit exists at a certain point. This definition uses two variables:

• Epsilon (\(\epsilon\)): Represents how close \(f(x)\) must be to the limit \(L\).
• Delta (\(\delta\)): Represents the allowed distance from \(c\) for \(x\) while ensuring \(f(x)\) stays within the proximity (determined by \(\epsilon\)) of \(L\).

The definition asserts that for every number \(\epsilon > 0\), there exists a number \(\delta > 0\) such that if \(0 < |x - c| < \delta\), then \(|f(x) - L| < \epsilon\). This theorem is a rigorous approach in calculus, ensuring that functions behave predictably as they approach certain points.

Finding Delta

Finding \(\delta\) is a crucial part of using the epsilon-delta definition in practice. The goal is to determine a \(\delta\) that works for a given \(\epsilon\) and a function \(f(x)\), ensuring \(|f(x) - L| < \epsilon\) for all \(x\) within the specified interval. Here's how you can proceed:

• First, express the inequality \(|f(x) - L| < \epsilon\).
• Solve for \(x\) to get bounds on \(|x - c|\).
• Identify the interval where \(x\) values satisfy the inequality.
• Calculate \(\delta\) as the smallest distance from the point \(c\) to the boundaries of this interval.

Using this process helps to establish a clear and workable interval where the conditions of the epsilon-delta definition are valid.

Limit of a Function

In calculus, finding the limit of a function is about determining how a function behaves as the variable approaches a certain value. The limit \(L\) is essentially the value that \(f(x)\) gets closer to as \(x\) approaches \(c\). Understanding limits involves:

• Identifying the value \(c\) at which \(x\) gets closer.
• Determining the value \(L\) that \(f(x)\) is approaching.
• Ensuring the function approaches \(L\) by being within any given proximity \(\epsilon\).

By mastering the concept of limits, students can understand key calculus topics like continuity, derivatives, and integrals, which all rely on the foundational idea of limits. In our example, finding \(\delta\) provides insight into how tightly the function adheres to the predicted behavior.