Problem 13 - Calculus: Early Transcendentals

Question

(a) Find a number \( \delta \) such that if \( | x - 2 | < \delta \), then \( | 4x - 8 | < \varepsilon \), where \( \varepsilon = 0.1 \). (b) Repeat part (a) with \( \varepsilon = 0.01 \).

Step-by-Step Solution

Answer

(a) \( \delta = 0.025 \); (b) \( \delta = 0.0025 \).

Step 1: Understand the problem

We want to find a number \( \delta \) such that if the distance between \( x \) and 2 is less than \( \delta \), then the expression \( 4x - 8 \) is within \( \varepsilon \) units of 0.

Step 2: Simplify the expression

Notice that \( 4x - 8 = 4(x - 2) \). We want \( |4(x - 2)| < \varepsilon \), which leads to \( 4|x - 2| < \varepsilon \).

Step 3: Solve for \( |x - 2| \)

From the equation \( 4|x - 2| < \varepsilon \), we can solve for \(|x - 2|\) by dividing both sides by 4. This gives us \( |x - 2| < \frac{\varepsilon}{4} \).

Step 4: Apply the given \( \varepsilon \) values

For part (a), \( \varepsilon = 0.1 \), so \( |x - 2| < \frac{0.1}{4} = 0.025 \). Therefore, \( \delta = 0.025 \). For part (b), \( \varepsilon = 0.01 \), so \( |x - 2| < \frac{0.01}{4} = 0.0025 \). Therefore, \( \delta = 0.0025 \).

Key Concepts

Epsilon-Delta Definition of a Limit

The epsilon-delta definition is a formal way to describe limits in calculus. It might sound a bit complex at first, but with careful steps, it becomes quite manageable.
The basic idea is to express how a function behaves as it approaches a certain point. Here's how it works:

• "Epsilon (\( \varepsilon \)", often pronounced as "epsilon") refers to how close we want the function value to be to the limit.
• "Delta (\( \delta \)") refers to how close we need to be to the point from the domain to achieve that function closeness.

For the exercise you're working on, this means finding a small range around the point 2 (given as \( x = 2 \)") so that the entire expression \( 4x - 8 \) stays within a given "epsilon-close" range from 0.
To determine how small we need our delta, we look at a simplified version of the equation, like we did with \( 4(x - 2) \), since it's in a form where we can easily decide the relationship between delta and epsilon.

Inequalities in Calculus

Inequalities come in handy when we're applying the epsilon-delta definition. They help us to express the condition:"whenever \( |x - L| < \delta \)", then "\( |f(x) - L| < \epsilon \)".
Here's a breakdown of how inequalities are used in this context:

• Identify the inequality: Start with an expression such as \(|4x - 8| < \varepsilon\).
• Simplify the expression: Express this inequality in a more digestible form like \(4|x - 2| < \varepsilon\)
• Solve for the domain: Rearrange it to make \(|x - 2|\) the center of attention, resulting in \( |x - 2| < \frac{\varepsilon}{4} \).

This process allows us to translate a complex function behavior around a point into something tangible and solve through known steps to define our delta.

Problem Solving in Calculus

Problem solving in calculus isn't just about crunching numbers; it's about understanding and applying concepts methodically.
Here’s how you can tackle problems like these step-by-step:

• Understand the problem: Clearly figure out what's being asked. Know the relationship between the variables and any conditions given.
• Simplify the problem: Break the expression into simpler forms. This often involves factoring or recognizing patterns, like we did with \( 4(x - 2) \).
• Apply concepts accurately: Use the definitions and properties you've learned, such as the epsilon-delta definition, to translate the problem into something solvable.
• Compute patiently: Carry out computations with care and check each step. Errors often happen with small arithmetic, especially with decimals like 0.1 and 0.01.

In our exercise, applying these steps provided precise solutions for different epsilon values. This structured approach is great for tackling many problems in calculus, ensuring accuracy and clarity in finding solutions.