Problem 29 - Calculus: Early Transcendentals

Question

Prove the statement using the \( \varepsilon \), \( \delta \) definition of a limit. \( \displaystyle \lim_{x \to 2} (x^2 - 4x + 5) = 1 \)

Step-by-Step Solution

Answer

Choose \( \delta = \sqrt{\varepsilon} \) to prove the limit.

Step 1: Understanding the Definition

The formal definition of a limit states that for every number \( \varepsilon > 0 \) there exists a number \( \delta > 0 \) such that if \( 0 < |x - 2| < \delta \), then \( |f(x) - L| < \varepsilon \), where for this function, \( f(x) = x^2 - 4x + 5 \) and \( L = 1 \).

Step 2: Compute the Function Value at the Limit Point

Substitute \( x = 2 \) into the function: \( f(2) = 2^2 - 4 \times 2 + 5 = 4 - 8 + 5 = 1 \). This confirms \( L = 1 \).

Step 3: Simplify the Expression \( |f(x) - L| \)

Consider the difference \( |f(x) - 1| = |x^2 - 4x + 5 - 1| = |x^2 - 4x + 4| = |(x - 2)^2| \).

Step 4: Relate \( |f(x) - 1| \) to \( |x - 2| \)

Since \( |(x - 2)^2| = |x - 2|^2 \), the condition \( |(x - 2)^2| < \varepsilon \) can be rearranged to \( |x - 2| < \sqrt{\varepsilon} \).

Step 5: Choose \( \delta \) in terms of \( \varepsilon \)

To satisfy the \( \varepsilon, \delta \) definition, choose \( \delta = \sqrt{\varepsilon} \). Therefore, if \( 0 < |x - 2| < \delta \), then \( |x - 2|^2 < \varepsilon \), i.e., \( |f(x) - 1| < \varepsilon \).

Key Concepts

Limit Theorem

The concept of a limit is foundational in calculus. It helps to understand how a function behaves as the input approaches a certain value. In our exercise, we need to prove that the limit of a function, as x approaches 2, equals 1. This involves identifying how close the function's output is to the limit value and managing any discrepancies.

The limit theorem can be constructed using the epsilon-delta definition, which literally quantifies "closeness" using two parameters:

• Epsilon (\( \varepsilon \)): Indicates the distance you will allow the function's output to deviate from the limit.
• Delta (\( \delta \)): Manages how close the input must be to the point of interest (in this case, 2) to ensure the output is within the epsilon distance.

This mathematical language guarantees that, no matter how small the epsilon (\( \varepsilon \)), there exists a delta (\( \delta \)) ensuring the function stays within bounds. This precision permits the formal proof ensuring the behavior of the function around specific points.

Calculus Proof

To prove the limit using the epsilon-delta definition, we must rely on a structured approach — a calculus proof.

For the function \( f(x) = x^2 - 4x + 5 \) and \( L = 1 \), we need to establish that \( \lim_{x \to 2} (x^2 - 4x + 5) = 1 \).
Here's the proof broken down:

• First, substitute the point of interest (2) into the function and verify that the function evaluates to the limit value (1), confirming \( L = 1 \).
• Next, calculate \( |f(x) - 1| \) and simplify it. For our example, it reduces to \( |x-2|^2 \), enabling us to say \( |x - 2|^2 < \varepsilon \), which is derived by resolving the expression through factorization.
• Finally, decide on a delta (\( \delta \)), and for us, this happens to align neatly as \( \delta = \sqrt{\varepsilon} \) to satisfy the inequality \( |x - 2| < \sqrt{\varepsilon} \).

This series of logical transformations and simple arithmetic verifies that the function's output remains appropriately close to the limit whenever the input nears 2.

Continuity in Calculus

Understanding continuity is vital to analyzing limits since continuous functions inherently have more straightforward limit proofs.

A function is continuous at a point if the limit as x approaches a certain value equals the function's value at that point. In mathematical terms for our function, this means \( \lim_{x \to 2} (x^2 - 4x + 5) = f(2) \).
Ensuring continuity involves:

• Checking that the function is defined at the point (here, at x = 2).
• Verifying that both the left-hand and right-hand limits at x = 2 exist and are equal.
• Confirming that the value of the limit at x = 2 matches the function value \( f(2) = 1 \).

The polynomial nature of our function \( f(x) \) guarantees it is smooth and continuous everywhere, particularly at x = 2. This allows for the direct application of the epsilon-delta definition without additional complications arising in discontinuous situations.