Problem 53 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(\delta\) (depending on \(\mathrm{s}\) ) so that the given implication is true. $$ |x-5|<\delta \Rightarrow|3 x-15|<\varepsilon $$

Step-by-Step Solution

Answer

\( \delta = \frac{\varepsilon}{3} \)

Step 1: Understand the Problem

We need to find a value for \( \delta \) such that the given implication \( |x-5|<\delta \Rightarrow|3x-15|<\varepsilon \) is true for any given \( \varepsilon \). This typically involves expressing \( |3x-15| \) in terms of \(|x-5|\) and then choosing \( \delta \) in terms of \( \varepsilon \).

Step 2: Relate the Expressions

Observe that the expression \( |3x-15| \) can be rewritten as \( 3|x-5| \) because \( |3x-15| = 3|x-5| \). This simplifies our process of finding \( \delta \).

Step 3: Establish the Inequality

Substitute \( 3|x-5| \) into the inequality, giving us \( 3|x-5| < \varepsilon \). This shows that if \( 3|x-5| < \varepsilon \) is true, then the implication holds.

Step 4: Solve for \( \delta \)

We want \( |x-5|<\delta \) to imply that \( 3|x-5|<\varepsilon \). Since \( 3|x-5|=3\delta \), we set \( 3\delta<\varepsilon \). Solving for \( \delta \), we have \( \delta<\frac{\varepsilon}{3} \).

Step 5: Choose \( \delta \)

Therefore, choose \( \delta = \frac{\varepsilon}{3} \) to satisfy the condition that \( |x-5|<\delta \Rightarrow|3x-15|<\varepsilon \) for any arbitrary \( \varepsilon \).

Key Concepts

Epsilon-delta definition

The epsilon-delta definition is a fundamental concept in calculus used to define the limit of a function. Imagine you want to prove that a function approaches a specific number as the input gets closer to a certain point. The epsilon-delta definition formalizes this notion.

• Epsilon (\(\varepsilon\)) represents any small positive number that measures how close the function's value should get to the limit.
• Delta (\(\delta\)) is another small positive number that indicates how near the input should be to the point of interest, ensuring the function's value stays within the epsilon range from the limit.

In simple terms, given a target output, if we can find an input restriction that keeps the function's outputs within a certain boundary, we can use epsilon and delta to prove the existence of a limit.
In this exercise, by choosing \(\delta = \frac{\varepsilon}{3}\), we've determined how the input domain is narrowed to meet the output criteria set by epsilon.

Inequalities in calculus

In calculus, inequalities are used to compare the sizes or magnitudes of numbers and expressions. They play a vital role in understanding the behavior of functions, especially when dealing with limits and continuity.
Consider the inequality \(|x-5|<\delta \Rightarrow |3x-15|<\varepsilon\).

• This inequality compares the distance between numbers using absolute values, ensuring the function never exceeds certain bounds.
• Inequalities help establish conditions under which functions behave predictably, which is crucial in ensuring the existence and value of limits.

To solve such inequalities, it's essential to manipulate terms smartly, as shown in this problem, where we simplified \(|3x-15|\) to \(3|x-5|\). This simplification allows us to express the relationship cleanly and precisely.

Continuity and limits

Continuity is all about smoothness or the idea that a function does not suddenly "jump". For a function to be continuous at a point, it must be smooth at that point and its limits should exist.

• A limit is a value that a function approaches as the input approaches a certain point. It's like predicting where a car will be on the road a moment before it actually gets there.
• The epsilon-delta method is a way of rigorously proving these limits and thus proving continuity. It ensures our predictions are mathematically sound.

In our exercise, we're dealing with limits when solving for \(\delta\). The process ensures that as \(x\) gets infinitely close to 5, \(3x-15\) will get infinitely close to becoming less than any desired positive \(\varepsilon\), consolidating the idea of continuity at that point.

Problem-solving in calculus

Problem-solving in calculus often revolves around methods to find limits, derivatives, and integrals of functions.
In this problem, we encountered a classic "epsilon-delta" challenge.

• First, understanding the conditions: recognizing \(|3x-15|\) could be simplified using algebraic identities.
• Next, strategizing the solution: using algebraic manipulation to rearrange the problem into a workable inequality.
• Finally, determining practicality: selecting \(\delta = \frac{\varepsilon}{3}\) provides a concrete solution, ensuring that every condition in the implication is satisfied for any small \(\varepsilon\).

By methodically tackling such problems, you improve your calculus skills, fostering a deeper understanding through consistent practice and application of theoretical foundations.