Problem 21 - Calculus: Early Transcendentals

Question

Prove the statement using the \( \varepsilon \), \( \delta \) definition of a limit. \( \displaystyle \lim_{x \to 4}\frac{x^2 - 2x - 8}{x - 4} = 6 \)

Step-by-Step Solution

Answer

The limit is proven by satisfying the \( \varepsilon \), \( \delta \) definition with \( \delta = \varepsilon \).

Step 1: Understand the Limit Definition

The limit of a function \( f(x) \) as \( x \) approaches \( c \) is \( L \) if for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x-c| < \delta \), the condition \( |f(x) - L| < \varepsilon \) holds. Here, \( c = 4 \) and \( L = 6 \).

Step 2: Simplify the Function

The given function is \( \frac{x^2 - 2x - 8}{x - 4} \). Factor the numerator: \( x^2 - 2x - 8 = (x-4)(x+2) \). So the function simplifies to \( x+2 \) for \( x eq 4 \).

Step 3: Apply the Limit Definition

We need \( |(x+2) - 6| < \varepsilon \) whenever \( 0 < |x-4| < \delta \). Simplifying, \( |x-4| < \varepsilon \).

Step 4: Choose Delta

Since \( |x-4| < \varepsilon \) already satisfies our definition for any \( \varepsilon > 0 \), we can choose \( \delta = \varepsilon \).

Step 5: Conclude the Proof

Thus, for every \( \varepsilon > 0 \), choosing \( \delta = \varepsilon \) ensures that \( |f(x) - 6| < \varepsilon \) whenever \( 0 < |x-4| < \delta \). Therefore, by the definition of a limit, \( \lim_{x \to 4}\frac{x^2 - 2x - 8}{x - 4} = 6 \).

Key Concepts

Limit of a Function

The concept of a limit is one of the foundational ideas in calculus. Understanding limits is crucial for studying continuous functions, derivatives, and integrals. A limit describes the behavior of a function as the input approaches a particular value. In mathematical terms, the limit of a function \( f(x) \) as \( x \) approaches \( c \) is \( L \) if, for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x-c| < \delta \), the inequality \( |f(x) - L| < \varepsilon \) holds. This essentially means that as \( x \) gets arbitrarily close to \( c \), \( f(x) \) gets arbitrarily close to \( L \).
In our exercise, we examined \( \lim_{x \to 4}\frac{x^2 - 2x - 8}{x - 4} = 6 \). By simplifying this function and applying the epsilon-delta definition, we can rigorously demonstrate that the limit indeed equals 6 when \( x \) approaches 4.

Proof Techniques in Calculus

When it comes to proving limits, especially using the epsilon-delta definition, certain proof techniques become invaluable. The primary objective is to connect the difference \( |f(x) - L| \) with \( |x - c| \), ensuring that it becomes less than \( \varepsilon \).
Here's a systematic approach to prove a limit using epsilon-delta:

• Start by analyzing the function and its behavior as \( x \) approaches the limit point \( c \).
• Simplify the expression when possible to make the algebra manageable.
• Identify a suitable \( \delta \) that allows the condition \( |f(x) - L| < \varepsilon \) to hold for all \( x \) within the delta range.

For the given function \( \frac{x^2 - 2x - 8}{x - 4} \), we simplified it to \( x+2 \) for all \( x eq 4 \), making it easier to see how the limit approaches 6 as \( x \) approaches 4. By comparing \( |(x+2)-6| \) to \( |x-4| \), we established that \( \delta = \varepsilon \) was a valid choice, effectively completing the proof.

Factoring in Algebra

Factoring is a powerful algebraic tool used to simplify expressions, solve equations, and in calculus, to find limits when indeterminate forms like zero over zero are encountered. In simple terms, factoring splits an expression into a product of simpler expressions. Skillful factoring can make seemingly complex math much more accessible.
In the exercise, we were tasked with handling \( \frac{x^2 - 2x - 8}{x - 4} \). The numerator \( x^2 - 2x - 8 \) needed factoring to eliminate the indeterminate form. We expressed the quadratic as \( (x - 4)(x + 2) \). This factoring step was crucial as it allowed us to cancel the \( x - 4 \) term, revealing the simplified form \( x + 2 \). Once in this form, it is straightforward to prove the limit using the epsilon-delta definition.
Mastery of factoring not only aids in evaluating limits but also strengthens overall algebra skills, proving invaluable across various mathematical contexts.