Problem 32 - Calculus: Early Transcendentals

Question

Prove the statement using the \( \varepsilon \), \( \delta \) definition of a limit. \( \displaystyle \lim_{x \to 2} x^3 = 8 \)

Step-by-Step Solution

Answer

Choose \(\delta = \min(1, \frac{\varepsilon}{19})\) to prove the limit.

Step 1 - Understand the definition

The goal is to prove that \(\lim_{{x \to 2}} x^3 = 8\) using the \(\varepsilon\), \(\delta\) definition of a limit. Recall that for \(\lim_{{x \to a}} f(x) = L\), for every \(\varepsilon > 0\) there exists a \(\delta > 0\) such that if \(0 < |x-a| < \delta\), then \(|f(x) - L| < \varepsilon\). In our case, \(a = 2\), \(f(x) = x^3\), and \(L = 8\).

Step 2 - Find expression for |f(x) - L|

We want to ensure that \(|x^3 - 8| < \varepsilon\). First, note that \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\). So, we need \(|(x - 2)(x^2 + 2x + 4)| < \varepsilon\).

Step 3 - Approximate bounds for x

We can assume \(x\) is close to \(2\). Let's choose \(\delta = 1\) for our approximation, meaning \(1 < x < 3\). This ensures \(|x - 2| < \delta\).

Step 4 - Estimate |x^2 + 2x + 4|

In the bounding interval \(1 < x < 3\), \(x^2 + 2x + 4\) varies between \(1^2 + 2(1) + 4 = 7\) and \(3^2 + 2(3) + 4 = 19\). Thus, \(|x^2 + 2x + 4| < 19\).

Step 5 - Establish condition for |(x - 2)(x^2 + 2x + 4)|

To ensure \(|(x - 2)(x^2 + 2x + 4)| < \varepsilon\), recognize that \(|x - 2| \times 19 < \varepsilon\). This means \(|x - 2| < \frac{\varepsilon}{19}\).

Step 6 - Choose \(\delta\) based on \(\varepsilon\)

Based on Step 5, we choose \(\delta = \min(1, \frac{\varepsilon}{19})\). This \(\delta\) guarantees that whenever \(|x - 2| < \delta\), the original condition \(|x^3 - 8| < \varepsilon\) is satisfied.

Key Concepts

Epsilon-Delta Definition

The \( \varepsilon\), \( \delta\) definition of a limit is foundational in calculus for proving the limit of a function as the input approaches a certain value. In simple terms, it bridges the concept of approaching a boundary without necessarily touching it. The definition states that for any arbitrary level of closeness \( \varepsilon > 0\), we can find a small enough region (\( \delta > 0\)) such that if a variable \( x\) is within \( \delta\) units of \( a\) (but not equal to \( a\)), then the function \( f(x)\) will be within \( \varepsilon\) units of the limit \( L\).
\( \)Here's how it unfolds:

• Start with \( \lim_{x o a} f(x) = L\).
• For every \( \varepsilon > 0\), there should be a \( \delta > 0\).
• If \( 0 < |x-a| < \delta\), then it follows that \( |f(x) - L| < \varepsilon\).

This method helps in representing the precision of limits in mathematical terms, making it possible to rigorously prove limit statements.

Polynomial Functions

Polynomial functions are a fundamental class of functions in algebra and calculus. They are composed of variables and coefficients, involving operations like addition, subtraction, multiplication, and non-negative integer exponents. A simple example is \( x^3\), which was the core function in our original exercise. Polynomials:

• Are generally easy to work with due to their predictable patterns.
• Have derivatives that are also polynomial functions, simplifying calculus operations.
• Feature roots (or zeros) determined by setting the polynomial to zero.

The limit exercise utilizes the expression \( x^3 - 8 = (x - 2)(x^2 + 2x + 4)\). This factorization is key for leveraging the properties of polynomials, specifically allowing us to understand how changes in \( x\) impact the function \( x^3\) around the point where \( x o 2\).
\( \)Polynomials, especially those of lower degrees like quadratic or cubic, make for simpler demonstrations of limit proofs due to their stable nature across real numbers.

Continuity in Calculus

Continuity is a crucial concept in calculus, indicating that a function has no abrupt jumps, breaks, or holes at a point. A function \( f(x)\) is continuous at a point \( a\) if the limit of \( f(x)\) as \( x o a\) equals \( f(a)\). Simply put, you can draw the function's graph around \( a\) without lifting your pencil.
\( \)

• A polynomial function like \( x^3\) is continuous across its entire domain due to its simple, smooth nature.
• At any point of \( x\), especially those involved in limit proofs, continuity ensures that manipulation and analysis remain straightforward.
• The \( \varepsilon\), \( \delta\) approach supports proving continuity by showing that as \( x\) approaches a point, the function's behavior remains predictable.

In the original exercise, demonstrating the limit at \( x = 2\) showcases continuity. By establishing that \( \lim_{x o 2} x^3 = 8\), we're affirming that there are no disruptions at \( x = 2\), proving \( x^3\)'s consistency and fluidity at that point.