Problem 14 - Calculus Early Transcendentals

Question

Given that \(\lim _{x \rightarrow 2}(5 x-7)=3,\) illustrate Definition 2 by finding values of \(\delta\) that correspond to \(\varepsilon=0.1, \varepsilon=0.05\) and \(\varepsilon=0.01\)

Step-by-Step Solution

Answer

\(\delta = 0.02\) for \(\varepsilon = 0.1\), \(\delta = 0.01\) for \(\varepsilon = 0.05\), \(\delta = 0.002\) for \(\varepsilon = 0.01\).

Step 1: Understand the Limit Definition

The definition For every there exists a (delta) such that if then indicates we want |f(x) - L| < \epsilon whenever |x - c| < \delta. Given: \( \lim_{x \to 2}(5x - 7) = 3 \). This means \( f(x) = 5x - 7 \), L = 3, and c = 2.

Step 2: Set up the Inequality for \(\varepsilon\)

We need \(|5x - 7 - 3| < \varepsilon\), which simplifies to \(|5x - 10| < \varepsilon\). This can be rewritten as \(5|x - 2| < \varepsilon\).

Step 3: Solve for \(\delta\)

Dividing through by 5 gives \(|x - 2| < \frac{\varepsilon}{5}\). This shows that \(\delta = \frac{\varepsilon}{5}\).

Step 4: Calculate \(\delta\) for \(\varepsilon = 0.1\)

Use \( \delta = \frac{\varepsilon}{5} \). For \(\varepsilon = 0.1\), \(\delta = \frac{0.1}{5} = 0.02\).

Step 5: Calculate \(\delta\) for \(\varepsilon = 0.05\)

For \(\varepsilon = 0.05\), use the formula \(\delta = \frac{0.05}{5} = 0.01\).

Step 6: Calculate \(\delta\) for \(\varepsilon = 0.01\)

For \(\varepsilon = 0.01\), use the formula \(\delta = \frac{0.01}{5} = 0.002\).

Key Concepts

Epsilon-Delta Definition

The concept of limits is central to calculus, and the epsilon-delta definition is the rigorous mathematical way to express this idea. Essentially, this definition formalizes what it means for a function to approach a certain value (the limit) as the input gets closer to a specific point. Imagine you have a function, and you're trying to see how it behaves near some point. The epsilon-delta definition helps you figure this out precisely.

Here's how it works:

• We begin with a number, \( L \), which is the limit we suspect our function will approach.
• For any small quantity \( \varepsilon > 0 \) ("epsilon"), which represents how close we want our function's value \( f(x) \) to be to \( L \), there is a corresponding small distance \( \delta > 0 \) ("delta").
• If \(|x - c| < \delta\) is true, then it must also be true that \(|f(x) - L| < \varepsilon\). This ensures that as we get closer to \( c \), the function \( f(x) \) gets arbitrarily close to \( L \).

Understanding this definition is crucial for solving limit problems, as it lays the groundwork for deep comprehension of how functions behave as inputs approach a specific value.

Limit Problems

Limit problems are the puzzles that calculus tries to solve. They show us how functions behave as they reach towards certain points. Consider the simple function \( f(x) = 5x - 7 \). To find out what happens as \( x \) approaches 2, you would determine the limit \( \lim_{x \to 2} (5x - 7) = 3 \). This means when \( x \) is very near 2, \( f(x) \) is very close to 3. Here's how to tackle limit problems with the epsilon-delta definition:

• Substitute the function into the inequality setup from the epsilon-delta method.
• In this example, set \(|5x - 7 - 3| < \varepsilon\), leading to \(|5x - 10| < \varepsilon\).
• Solving this gives \(5|x - 2| < \varepsilon\).

This process allows you to see how the function reacts as \( x \) gets closer to a particular point. It's about knowing the boundaries and behavior of the function, which is foundational in calculus.

Calculating Delta

Once you understand the setup of limit problems, calculating \( \delta \) becomes a manageable task. The main goal is to express \( \delta \) in terms of \( \varepsilon \) so you know how changes in one affect the other. Using our previous function \( f(x) = 5x - 7 \), and for any \( \varepsilon \), you'd perform these steps:

• From the inequality \(5|x - 2| < \varepsilon\), solve for \(|x - 2|\).
• Isolate \(x - 2\) to get \(|x - 2| < \frac{\varepsilon}{5}\).

This makes \( \delta = \frac{\varepsilon}{5} \), which expresses exactly how close \( x \) should be to 2 to keep \( f(x) \) within \( \varepsilon \) of the limit. Let's calculate \( \delta \) for different \( \varepsilon \) values:

• For \( \varepsilon = 0.1 \),\( \delta = \frac{0.1}{5} = 0.02 \).
• For \( \varepsilon = 0.05 \), \( \delta = \frac{0.05}{5} = 0.01 \).
• For \( \varepsilon = 0.01 \), \( \delta = \frac{0.01}{5} = 0.002 \).

Each \( \varepsilon \) value shows how much you can vary \( x \) and still have the function within the target range of the limit. This calculation gives you control over the precision of the function's limit, an essential tool in calculus.