Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

Give an \(\varepsilon-\delta\) proof of each limit fact. $$ \lim _{x \rightarrow 5} \frac{x^{2}-25}{x-5}=10 $$

Step-by-Step Solution

Answer

The limit is 10, proven using the \(\varepsilon-\delta\) method with \(\delta = \varepsilon\).

Step 1: Understand the Goal

Our goal is to prove that \( \lim _{x \rightarrow 5} \frac{x^2 - 25}{x-5} = 10 \) using the \( \varepsilon-\delta \) definition of a limit. This means we need to show that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 5| < \delta \), then \( \left| \frac{x^2 - 25}{x-5} - 10 \right| < \varepsilon \).

Step 2: Simplify the Function

Notice that the expression \( \frac{x^2 - 25}{x-5} \) can be simplified. Factor the numerator as \( (x-5)(x+5) \). So, \( \frac{x^2 - 25}{x-5} = x + 5 \) for \( x eq 5 \). Our function simplifies to \( f(x) = x + 5 \).

Step 3: Apply the \( \varepsilon-\delta \) Definition

We now need to prove that \( \lim_{x \rightarrow 5} (x + 5) = 10 \) using the \( \varepsilon-\delta \) criterion. For this, consider \( |(x+5) - 10| < \varepsilon \). Simplifying gives \( |x - 5| < \varepsilon \).

Step 4: Find \( \delta \) in Terms of \( \varepsilon \)

We need \( |x - 5| < \varepsilon \) to hold, which implies that we can choose our \( \delta \) to be exactly \( \varepsilon \). Therefore, for every \( \varepsilon > 0 \), let \( \delta = \varepsilon \).

Step 5: Conclude the Proof

Because for every \( \varepsilon > 0 \), choosing \( \delta = \varepsilon \) ensures that \( |x-5| < \varepsilon \), satisfying the \( \varepsilon-\delta \) definition of a limit. Hence, the limit is proven: \( \lim _{x \rightarrow 5} \frac{x^2-25}{x-5}=10 \).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta (\( \varepsilon-\delta \)) definition is a formal approach to prove limits in calculus. It describes the behavior of a function as it approaches a particular point. The essence of this definition is to prove how the output values of a function can be made arbitrarily close to a limit, by making the input values sufficiently close to a specific point. In simpler terms, given a function \( f(x) \) approaching a limit \( L \) as \( x \) approaches \( a \), this definition states:

• For every positive number \( \varepsilon \) (no matter how small), there's a corresponding positive number \( \delta \).
• If \( 0 < |x - a| < \delta, \) then \( |f(x) - L| < \varepsilon.\)

This framework ensures the function value is as close to \( L \) as desired, based on how close \( x \) is to \( a. \) This concept underpins the rigorous foundation of calculus, offering precise control over how functions behave near particular points.

Limit Proof

Proving limits using the \( \varepsilon-\delta \) definition is essential for validating how functions act when variables approach specific values. In our example, the task is to prove:\[\lim_{x \rightarrow 5} \frac{x^2 - 25}{x-5} = 10\]Given the function \( f(x) = \frac{x^2 - 25}{x-5}, \) the first step involves simplifying it to \( f(x) = x + 5 \) when \( x ot= 5. \) To proceed with the \( \varepsilon-\delta \) proof, you need to show that for all \( \varepsilon > 0, \) a corresponding \( \delta > 0\) can be determined so that \( |x - 5| < \delta \) implies:

• \( |f(x) - 10| < \varepsilon.\)
• Simplifying it results in \( |x + 5 - 10| < \varepsilon \), which further simplifies to \( |x - 5| < \varepsilon.\)
• Here, selecting \( \delta = \varepsilon \) accomplishes this requirement, concluding the proof.

This systematic approach assures us that the function approaches the limit exactly as defined, reinforcing the mathematical robustness of limit proofs.

Algebraic Simplification

Algebraic simplification is a crucial step in solving limits, especially when expressions seem complex initially. Our goal is to ease the process of evaluating limits by breaking down expressions into simpler forms. Consider the expression used in our problem:\[\frac{x^2 - 25}{x-5}\]To simplify, recognize this as a difference of squares:

• Factor the numerator: \( x^2 - 25 = (x-5)(x+5). \)
• Cancelling out \( x-5 \), we simplify the function to \( x + 5 \) wherever \( x eq 5.\)

This simplified form \( x + 5 \) is much easier to handle and analyze when evaluating limits. Simplifying expressions in this way is an essential skill, enabling clearer insights into the behavior of functions near specific points. It is a powerful algebraic tool that ensures a simpler, more intuitive path to finding limits, reducing potential for errors in evaluation.