Problem 30 - Calculus Early Transcendentals

Question

Prove the statement using the \(\varepsilon, \delta\) definition of limit. \(\lim _{x \rightarrow 3}\left(x^{2}+x-4\right)=8\)

Step-by-Step Solution

Answer

The limit is proven using the \( \varepsilon, \delta \) definition by showing \( \delta = \min(1, \varepsilon/8) \) works for any \( \varepsilon > 0 \).

Step 1: Understand the problem

We are tasked with proving that \( \lim _{x \to 3}(x^2 + x - 4) = 8 \) using the \( \varepsilon, \delta \) definition of a limit. This means for every \( \varepsilon > 0 \), we need a \( \delta > 0 \) such that whenever \( 0 < |x - 3| < \delta \), it follows that \( |(x^2 + x - 4) - 8| < \varepsilon \).

Step 2: Simplify the expression

First, simplify \( |(x^2 + x - 4) - 8| \) to make it easier to work with. The expression simplifies to:\[|(x^2 + x - 12)| = |x^2 + x - 12|\]

Step 3: Factor the expression

Attempt to factor the quadratic \( x^2 + x - 12 \). Factoring, we get:\[|x^2 + x - 12| = |(x - 3)(x + 4)|\]

Step 4: Set up \( \varepsilon \) condition

We want to find \( \delta \) such that the expression \( |(x - 3)(x + 4)| < \varepsilon \). Since we have \( 0 < |x - 3| < \delta \), we focus on bounding \( |x + 4| \).

Step 5: Assume a bound for \( x \)

Assume \( |x - 3| < 1 \), which implies \( 2 < x < 4 \). Consequently, \(6 < x + 4 < 8\). Thus, \( |x + 4| < 8 \).

Step 6: Relate \( \varepsilon \) and \( \delta \)

From \( |(x - 3)(x + 4)| < \varepsilon \), and \( |x + 4| < 8 \), we derive:\[|(x - 3)||x + 4| < 8|x - 3| < \varepsilon\]Thus, choose \( \delta = \min(1, \varepsilon/8) \).

Step 7: Conclusion of proof

For \( \varepsilon > 0 \), if \( \delta = \min(1, \varepsilon/8) \) and \( 0 < |x - 3| < \delta \), then \( |(x - 3)(x + 4)| < \varepsilon \). Hence, the limit is proven according to the \( \varepsilon, \delta \) definition.

Key Concepts

Limit of a Function

In calculus, understanding the concept of a limit is fundamental. A limit helps describe the behavior of a function as its input approaches a specific value. For instance, in our exercise, we are interested in how the function \( f(x) = x^2 + x - 4 \) behaves as \( x \) approaches 3. Mathematically, we express this as \( \lim_{x \to 3}(x^2 + x - 4) = 8 \).
Simply put, this notation means that as \( x \) gets closer and closer to 3, the output of the function \( f(x) \) gets closer to 8. The limit doesn't necessarily imply that when \( x \) is exactly 3, the value of the function will be 8, but rather it suggests that approaching is the key consideration.
This idea is essential for further topics, such as continuous functions and derivatives, which build upon the concept of limits.

Delta-Epsilon Proof

The delta-epsilon definition of a limit is a precise mathematical way to define when the limit of a function equals a specific number. In our exercise, this proof requires us to find a \( \delta \) for any given \( \varepsilon \) such that if \( 0 < |x - 3| < \delta \), then \( |f(x) - 8| < \varepsilon \).
Here's how it works:

• \( \varepsilon \) represents how close we want \( f(x) \) to be to the limit \( L \), which is 8 in this case.
• \( \delta \) represents a range around \( x = 3 \) that keeps \( f(x) \) within \( \varepsilon \) of 8.

The aim is to ensure that for every \( \varepsilon \), no matter how small, there will exist a corresponding \( \delta \) ensuring the condition holds. Thus, this approach provides a systematic method to rigorously prove the behaviors of functions around given points.

Quadratic Expressions

A quadratic expression is a polynomial of degree 2, commonly written in the form \( ax^2 + bx + c \). It is characterized by having its highest exponent of the variable, \( x \), as 2. In our exercise, the expression \( x^2 + x - 12 \) fits this form.
Quadratic expressions appear frequently in various mathematical problems, including solving equations, analyzing parabolas, and optimizing values in calculus. Understanding how to manipulate and transform these expressions is crucial for deeper mathematical study.
Key features of quadratic expressions include:

• The coefficients \( a \), \( b \), and \( c \) which determine the shape and position of the parabola when graphed.
• The ability to be factored or solved using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• The presence of roots which are solutions to the equation when set to zero.

Mastering these characteristics enables a strong foundation for understanding more complex functions and calculus concepts.

Factoring Polynomials

Factoring polynomials is a powerful technique used to simplify quadratic expressions and polynomial functions. The process involves breaking a polynomial into a product of its factors, which are simpler expressions that multiply to give the original polynomial. In our problem, \( x^2 + x - 12 \) is factored into \( (x - 3)(x + 4) \).
Here's how factoring helps:

• It simplifies expressions, making them easier to work with, such as solving equations or evaluating limits.
• It reveals important properties of the polynomial, like its roots, which are solutions to the equation when set to zero.
• It is essential for completing the square, another method used to solve quadratic equations.

Factors like \( (x - 3)(x + 4) \) indicate points where the expression equals zero, which is essential for understanding the graph's intercepts and behavior. Thus, factoring is a foundational tool in algebra and calculus, providing insights necessary for problem-solving and mathematical reasoning.