Problem 55 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(\delta\) (depending on \(\varepsilon\) ) so that the given implication is true. $$ |x+6|<\delta \Rightarrow|6 x+36|<\varepsilon $$

Step-by-Step Solution

Answer

Set \(\delta = \frac{\varepsilon}{6}\).

Step 1: Analyze the Given Expression

We start by examining the implications we want to establish: \(|x+6|<\delta \Rightarrow |6x+36|<\varepsilon\). Our task is to find \(\delta\) that depends on \(\varepsilon\) such that this implication holds.

Step 2: Simplify the Right Side

Notice, the expression \(|6x+36|\) can be rewritten as \(6|x+6|\) because \(6x+36 = 6(x+6)\). So, the implication is stated as: \(|x+6|<\delta \Rightarrow 6|x+6|<\varepsilon\).

Step 3: Relate the Conditions

For the implication \(6|x+6|<\varepsilon\) to hold true, we need to relate \(\delta\) to \(\varepsilon\). By dividing \(\varepsilon\) by 6, we can set \(6|x+6| < \varepsilon\) to match the left side condition. Therefore, set \(\delta = \frac{\varepsilon}{6}\).

Step 4: Check the Validity

If \(|x+6| < \delta = \frac{\varepsilon}{6}\), then multiplying both sides by 6 gives \(6|x+6| < 6 \times \frac{\varepsilon}{6} = \varepsilon\). This satisfies the requirement that \(|6x+36| < \varepsilon\). Therefore, our choice for \(\delta\) is valid.

Key Concepts

Implication

In mathematics, an implication is a logical relationship between two statements or conditions. In this exercise, the implication is expressed as \(|x+6|<\delta \Rightarrow |6x+36|<\varepsilon\). This means that if the first condition, \(|x+6|<\delta\), is true, then the second condition, \(|6x+36|<\varepsilon\), must also be true. This is a typical scenario in calculus where we are asked to find a connection between two inequalities, which is crucial in proving limits and continuity.

To work with implications efficiently, it is often useful to express the conditions in a form that shows how one depends on the other. In this case, simplifying both sides allowed us to see that if we determine \(\delta\) properly in terms of \(\varepsilon\), the implication will hold.

When dealing with implications in mathematics, it's helpful to:

• Identify clearly the hypothesis and conclusion.
• Use known mathematical properties to simplify expressions.
• Rearrange or process the expressions to make the dependency clearer.

Inequality

An inequality is a mathematical statement that indicates that two expressions are not equal, where one is either greater or less than the other. In our problem, we deal with the inequalities \(|x+6|<\delta\) and \(|6x+36|<\varepsilon\). These inequalities describe a range within which the variables can vary.

The inequality \(|x+6|<\delta\) suggests that \(x\) is kept close to -6, within a distance of \(\delta\). Similarly, \(|6x+36|<\varepsilon\) constrains the expression \(6x+36\) to be within a distance of \(\varepsilon\) from 0.

In solving problems involving inequalities, it's helpful to:

• Simplify expressions using factorization or division, like rewriting \(|6x+36|\) as \(6|x+6|\).
• Adjust related values to match, as we equated \(6|x+6|\) to \(\varepsilon\) by dividing \(\varepsilon\) by 6.

Epsilon-Delta Definition

The epsilon-delta definition is a fundamental concept in calculus, used to rigorously define the limit of a function. It is expressed in terms of two parameters: epsilon (\(\varepsilon\)) and delta (\(\delta\)), which represent distances or tolerances.

In practical terms, \(\varepsilon\) is the allowable error in the function's output, and \(\delta\) is the allowable error in the input. In our exercise, to ensure \(|6x+36|<\varepsilon\), we need \(|x+6|<\delta\), aligning with the definition of limits.

When using the epsilon-delta definition:

• Determine how small \(\delta\) must be for a given \(\varepsilon\), ensuring the function stays within the epsilon limit.
• Understand that this approach is about making the output of a function as close as desired to a limit value by adjusting the input.
• Reformulate expressions to clearly see how changes in one lead to changes in another.

Using this method allows us to make precise statements about continuity and limits, as we find \(\delta = \frac{\varepsilon}{6}\) to satisfy the problem's requirements.