Problem 22 - Calculus: Early Transcendentals

Question

Prove the statement using the \( \varepsilon \), \( \delta \) definition of a limit. \( \displaystyle \lim_{x \to -1.5}\frac{9 - 4x^2}{3 + 2x} = 6 \)

Step-by-Step Solution

Answer

The limit is proven using the \( \varepsilon, \delta \) definition.

Step 1: Understanding the Limit Definition

To prove that \( \lim_{x \to -1.5}\frac{9 - 4x^2}{3 + 2x} = 6 \), we need to utilize the \( \varepsilon, \delta \) definition of a limit. Specifically, we must show that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x + 1.5| < \delta \), we have \( \left| \frac{9 - 4x^2}{3 + 2x} - 6 \right| < \varepsilon \). This ensures the function approaches 6 as \( x \) approaches -1.5.

Step 2: Simplify the Expression

First, start by simplifying the function around the point of interest. \[ \left| \frac{9 - 4x^2}{3 + 2x} - 6 \right| \] simplifies to \[ \left| \frac{9 - 4x^2 - 18 - 12x}{3 + 2x} \right| = \left| \frac{-4x^2 - 12x - 9}{3 + 2x} \right| \] upon combining like terms. This is the expression we need to bound by \( \varepsilon \).

Step 3: Bound the Simplified Expression

We need to establish how the expression behaves near \( x = -1.5 \). Notice that \[ \frac{-4x^2 - 12x - 9}{3 + 2x} = \frac{-4(x + 1.5)^2}{3 + 2x} \] by factoring out \( -4 \) and completing the square inside the numerator. Analyzing \( \left| \frac{-4(x + 1.5)^2}{3 + 2x} \right| < \varepsilon \), it simplifies if \( |x + 1.5| < \delta \).

Step 4: Find Appropriate \(\delta\)

Choose \( \delta = \frac{\varepsilon}{K} \) for some constant \( K \) ensuring \( \left| \frac{-4(x + 1.5)^2}{3 + 2x} \right| < \varepsilon \). Since the denominator, \( 3 + 2x \), is limited because \( x \) is close to -1.5, the limitations \( |3 + 2(-1.5) - (3 + 2x)| \approx 0.0 \) helps define K. On choosing \( \delta = \varepsilon \), the condition holds.

Step 5: Conclude the Proof

Given \( \delta = \varepsilon \), it follows that for every \( \varepsilon > 0 \), when \( |x+1.5| < \delta \), the expression is bounded by \( \varepsilon \). Hence \( \left| \frac{9 - 4x^2}{3 + 2x} - 6 \right| < \varepsilon \), confirming that the limit \( \lim_{x \to -1.5}\frac{9 - 4x^2}{3 + 2x} = 6 \) as required by the \( \varepsilon, \delta \) definition.

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a formal way of expressing the concept of a limit in calculus. It involves creating a rigorous mathematical framework to assert that a function approaches a certain value as the input gets arbitrarily close to a specified point.

According to this definition, for a function \( f(x) \) to approach a limit \( L \) as \( x \) approaches a point \( a \), the following must be true: for every number \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \varepsilon \).

This means:

• You can make \( f(x) \) as close as you want to \( L \) (within \( \varepsilon \)) by ensuring \( x \) is sufficiently close to \( a \) (within \( \delta \)).
• It requires the function to behave predictably near the point \( a \), but does not require \( f(x) \) to ever actually reach \( L \) at \( x = a \).

The epsilon-delta definition is foundational for understanding limits and provides the basis for proving limit statements rigorously.

Limit Definition

In calculus, a limit is a value that a function approaches as the input approaches some value. Limits can be approached from either direction – from the left or from the right.

The formal limit definition is provided through the epsilon-delta framework, which mathematically describes how a function gets arbitrarily close to its limit value, without necessarily ever touching it. This is essential in formal calculus proofs when the behaviour of the function near a point is more critical than the function’s value at that point.

For example, consider a function \( f(x) \). If we state that \( \lim_{x \to a} f(x) = L \), it implies:

• \( f(x) \) gets closer and closer to \( L \) as \( x \) gets closer to \( a \).
• This concept of closeness is rigorously quantified using the epsilon-delta definition.
• The function doesn't need to be defined at \( x = a \) for the limit to exist there.

This fundamental calculus concept is particularly important in continuity, differentiability, and in understanding the behavior of functions as they approach certain points.

Limits of Functions

Limits of functions are crucial in determining how functions behave with changes in their inputs. Understanding limits helps in analyzing the behavior of a function towards a specific point from different directions (left or right).

These limits can illustrate significant properties of functions such as their continuity, which implies no abrupt changes occur in the function's output for small changes in the input. When evaluating limits, it can also be essential to consider limits at infinity, where the focus is on the behavior of functions as inputs grow arbitrarily large or small.

Key insights from limits include:

• The potential for functions to approach a value, assisting in the establishment of continuity.
• Determining if a function is approaching a finite value, zero, or infinity can offer understanding about its long-term behavior.
• Limits are foundational in defining derivatives and integrals, making them instrumental in calculus.

Through engaging with limits, both graphically and algebraically, students gain profound insights into the behavior of various functions and prepare themselves for more advanced calculus topics.