Problem 53 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

\( \delta = \frac{\varepsilon}{3} \) ensures the implication is true.

Step 1: Understand the Implication

We need to show that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that \( |x-5| < \delta \) implies \( |3x-15| < \varepsilon \).

Step 2: Simplify the Right-hand Side

Notice that \( 3x - 15 = 3(x - 5) \). Therefore, \( |3x - 15| = 3 |x - 5| \).

Step 3: Use Simplified Expression

The condition \( |3x - 15| < \varepsilon \) becomes \( 3|x - 5| < \varepsilon \).

Step 4: Solve for \( |x - 5| \) in terms of \( \varepsilon \)

From \( 3|x - 5| < \varepsilon \), we solve for \( |x - 5| \): \( |x - 5| < \frac{\varepsilon}{3} \).

Step 5: Define \( \delta \)

To ensure the implication is true, choose \( \delta = \frac{\varepsilon}{3} \). This means if \( |x - 5| < \delta \), then \( |3x - 15| < \varepsilon \).

Key Concepts

Limit

In calculus, the idea of a limit is fundamental. It helps us understand how functions behave as they approach certain points. A limit describes the value that a function, or a sequence, "approaches" as the input approaches some point. Here, we're taking the limit of a function as the variable gets infinitely close to a point.The notation \( \lim_{{x \to a}} f(x) = L \) means that as \( x \) approaches \( a \), the function \( f(x) \) approaches the value \( L \). Limits are crucial for defining derivatives and integrals as well.In our problem, we are finding \( \delta \) in terms of \( \varepsilon \) to ensure that the distance of \( x \) from 5 remains small enough to keep the function's output close, under a defined threshold. This ensures the change in the function remains controlled and within specific bounds, illustrating a core idea associated with limits.

Continuity

Continuity in mathematics means that a function does not have any jumps, breaks, or holes. For a function \( f(x) \) to be continuous at a point \( x = a \), three conditions must be satisfied:

• \( f(a) \) is defined.
• The limit \( \lim_{{x \to a}} f(x) \) exists.
• \( \lim_{{x \to a}} f(x) = f(a) \).

The epsilon-delta definition of a limit also provides the groundwork for continuity. If you can find a delta for every epsilon around a point, it means the function can be made arbitrarily close to the function value by choosing delta small enough. This concept is important in our context, as the problem explores how close inputs \( x \) need to be to 5 to ensure closeness of the function output, assuring continuity.

Epsilon-delta proof

An epsilon-delta proof is a formal way to prove statements about limits. It shows that the output of a function can get arbitrarily close to a certain value (\( L \)) by making the input close enough to a target point (\( a \)). The task of such proofs is to find \( \delta \) for every given \( \varepsilon > 0 \) to guarantee: \[ |x-a| < \delta \Rightarrow |f(x)-L| < \varepsilon \]In our given problem, we are finding such a relationship, where \( |x-5| < \delta \) must ensure \( |3x-15| < \varepsilon \). By manipulating the equation, as shown in the solution, we determine that \( \delta = \frac{\varepsilon}{3} \) is suitable. This careful arrangement illustrates the strength of epsilon-delta proofs in ensuring that limits are met precisely, demonstrating its power in calculus.

Absolute value inequalities

Absolute value inequalities involve expressions where we're interested in the distance from zero. They have the format \( |a| < b \), which means that the value \( a \) is between \(-b\) and \( b \).In the context of the given problem, the absolute value inequality \( |x-5| < \delta \) focuses on how far \( x \) is from 5. Solving these inequalities often involves breaking them into two linear inequalities, handling them as \( -b < a < b \).For the problem, converting \( |3x-15| < \varepsilon \) to \( 3|x-5| < \varepsilon \) helped simplify finding \( \delta \) as \( \frac{\varepsilon}{3} \). Mastery of absolute value inequalities is crucial in calculus for dealing with functions and limits with precision.