Problem 23 - 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. $$ f(x)=x^{2}, \quad L=4, \quad c=-2, \quad \epsilon=0.5 $$

Step-by-Step Solution

Answer

Open interval: \((-\sqrt{4.5}, -\sqrt{3.5})\). \(\delta = 0.121\).

Step 1: Understanding the Inequality

We need to find an interval around \( c = -2 \) such that \( |f(x) - L| < \epsilon \) holds, where \( f(x) = x^2 \), \( L = 4 \), and \( \epsilon = 0.5 \). The inequality becomes \( |x^2 - 4| < 0.5 \).

Step 2: Simplifying the Inequality

Rewrite \( |x^2 - 4| < 0.5 \) as a double inequality: \(-0.5 < x^2 - 4 < 0.5\). This simplifies to \(3.5 < x^2 < 4.5\). We need to find an open interval for \(x\) that satisfies this condition.

Step 3: Finding the Interval for x

Solve the inequalities: \(x^2 > 3.5\) and \(x^2 < 4.5\). For \(x^2 > 3.5\), \(|x| > \sqrt{3.5}\), which gives \(x > \sqrt{3.5}\) or \(x < -\sqrt{3.5}\). For \(x^2 < 4.5\), \(-\sqrt{4.5} < x < \sqrt{4.5}\). Thus, the interval for \(x\) is \((-\sqrt{4.5}, -\sqrt{3.5}) \cup (\sqrt{3.5}, \sqrt{4.5}) \).

Step 4: Focusing on the Interval around c = -2

Since \( c = -2 \), we only consider the negative part of the interval which contains \( -2 \), i.e., \(-(\sqrt{4.5}), -\sqrt{3.5})\).

Step 5: Finding \( \delta \)

To find \( \delta \), we need \(0 < |x + 2| < \delta\) implies \(x \in (-(\sqrt{4.5}), -\sqrt{3.5})\). The closest endpoint to \(-2\) in this interval is \(-\sqrt{4.5}\). Calculate \(\delta = |-2 + \sqrt{4.5}|\). Thus, \(\delta \approx |\sqrt{4.5} - 2|\).

Step 6: Conclusion for \( \delta \)

Compute \(\delta \): \(\sqrt{4.5} \approx 2.121\). Therefore, \(\delta \approx |2.121 - 2| = 0.121\). Thus, a suitable value for \(\delta\) is \(\delta = 0.121\).

Key Concepts

Inequalities

In this exercise, inequalities play a crucial role in understanding the limits and continuities of functions. An inequality expresses the relationship between two expressions that are not equal and is usually denoted using symbols such as \(<\), \(>\), \(\leq\), and \(\geq\). Here, we use the inequality \(|f(x) - L| < \epsilon\), which ensures that the values of the function \(f(x)\) remain within a specific range from \(L\).
For the given function \(f(x) = x^2\), \(L = 4\), and \(\epsilon = 0.5\), the inequality becomes \(|x^2 - 4| < 0.5\). This means that the values of \(x^2\) should lie between 3.5 and 4.5. We express this as the double inequality \(-0.5 < x^2 - 4 < 0.5\), which is simplified further to \(3.5 < x^2 < 4.5\).
Understanding how to manipulate and solve these inequalities is pivotal for determining open intervals and limits.

• Use inequalities to bound function values around a limit.
• Solve compound inequalities to find a valid range for \(x\).

Delta-Epsilon Definition of Limit

The delta-epsilon definition of limit is a formal definition used to grasp the concept of limits in calculus. It rigorously describes how close the function \(f(x)\) approaches a limit \(L\) as \(x\) approaches a value \(c\).
In simpler terms, this definition provides a method to make \(f(x)\) arbitrarily close to \(L\) by making \(x\) sufficiently close to \(c\). We denote this with two small positive numbers \(\delta\) and \(\epsilon\) , where \(0 < |x - c| < \delta\) guarantees that \(|f(x) - L| < \epsilon\).

• \(\delta\) is the maximum allowable distance of \(x\) from \(c\).
• \(\epsilon\) is the maximum allowable distance of \(f(x)\) from \(L\).

For our function \(f(x) = x^2\) with \(c = -2\), we find that \(\delta = |\sqrt{4.5} - 2| \approx 0.121\) ensures \(|x^2 - 4| < 0.5\) holds true when \(x\) is close to \(-2\).
This methodology helps to precisely determine how small \(\delta\) needs to be to achieve the desired closeness to \(L\).

Open Intervals

Open intervals involve a range of values where the endpoints are not included in the interval. They are denoted by parentheses, such as \((a, b)\), which includes all numbers between \(a\) and \(b\) but not \(a\) or \(b\) themselves.
In the context of this exercise, it was necessary to find an open interval around \(c = -2\) where the inequality \(|f(x) - L| < \epsilon\) is satisfied. We determined that \(f(x) = x^2\) lies in two separate open intervals \((\sqrt{3.5}, \sqrt{4.5})\) and \((-\sqrt{4.5}, -\sqrt{3.5})\).
Since \(c = -2\), we focus on the interval \((-\sqrt{4.5}, -\sqrt{3.5})\), which contains \(-2\). This way, the function meets the inequality condition within this open segment.

• Open intervals help determine where specific conditions hold true.
• Useful in defining neighborhoods around a particular point, such as \(-2\).

They are essential in understanding where limits behave consistently, ensuring the calculations stick to legitimate ranges of \(x\).