Problem 24 - Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The statement is true; the limit is always \( c \) for a constant function.

Step 1: Understanding the Limit Statement

The statement \( \lim_{x \to a} c = c \) means that as \( x \) approaches \( a \), the function \( f(x) = c \) approaches the value \( c \). Here, \( c \) is a constant function, meaning its value does not depend on \( x \).

Step 2: Recall the Epsilon-Delta Definition

The \( \varepsilon, \delta \) definition of a limit states that for any \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x - a| < \delta \), it follows that \( |f(x) - L| < \varepsilon \). In this problem, \( L = c \) and \( f(x) = c \).

Step 3: Setup the Epsilon-Delta Proof

We need to show that for every \( \varepsilon > 0 \), we can find a \( \delta > 0 \) such that \( |f(x) - c| < \varepsilon \) whenever \( 0 < |x - a| < \delta \). In this case, \( f(x) = c \), so \( |c - c| = 0 \).

Step 4: Picking the Delta Value

Since \( |c - c| = 0 \), which is always less than any \( \varepsilon > 0 \), we can choose any positive \( \delta \). In reality, \( \delta \) does not impact the condition because \( f(x) = c \) everywhere.

Step 5: Concluding the Proof

Given that \( |c - c| = 0 < \varepsilon \) is true for any \( \varepsilon > 0 \) and any positive \( \delta \), the condition required by the limit definition is satisfied. Therefore, \( \lim_{x \to a} c = c \) is proven by the \( \varepsilon, \delta \) definition.

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a fundamental concept for understanding limits in calculus. It provides a rigorous mathematical way to describe what it means for a function to approach a certain value. This definition states:

• For any positive number \( \varepsilon \), there exists another positive number \( \delta \).
• For every \( x \) within the interval defined by \( 0 < |x - a| < \delta \), the function value \( f(x) \) is within \( \varepsilon \) distance of \( L \), which is the limit we claim \( f(x) \) approaches.

In simple terms, as \( x \) gets closer to \( a \), \( f(x) \) gets arbitrarily close to \( L \). This precise formulation is essential for many higher-level concepts in calculus. It ensures we can reliably say a function behaves a certain way as it approaches a particular point, including determining whether a function is continuous or differentiable.

Constant Function

A constant function is one of the simplest types of functions you'll encounter in calculus. It is defined as a function that returns the same value no matter what is chosen for \( x \). Mathematically, if \( f(x) = c \) for some constant \( c \), then \( f(x) \) is a constant function.

• Its graph is a horizontal line on the Cartesian plane.
• Every point on the line corresponds to the same value \( c \), showing its constancy.

The significance of a constant function in limits is that for any \( x \), the function does not change, making proving limits fairly straightforward.
In the given problem, the function \( f(x) = c \) simplifies the process because no matter how \( x \) changes, the output remains constant. This means that it naturally satisfies the conditions required by the epsilon-delta definition without complex adjustments. A constant function is crucial for building understanding, as it is foundational in learning how limits interact with different types of functions.

Proof in Calculus

Proofs are at the heart of understanding mathematics deeply. In calculus, proofs allow us to show that certain intuitions or results hold true for all cases under consideration. The step-by-step proof of a limit statement using the epsilon-delta definition demonstrates meticulous logical reasoning.

• Start by understanding the limit statement, translating it into a form suitable for the epsilon-delta approach.
• Use the epsilon-delta condition to find appropriate values of \( \delta \) for every \( \varepsilon \).
• Conclude by showing these values adhere to the necessary conditions, thereby proving the limit exists.

In the given task, because \( f(x) = c \) is constant, the proof simplifies beautifully. Since \( |c - c| = 0 \) is always zero, it will always be less than any positive \( \varepsilon \). Thus, any positive \( \delta \) meets the required condition, making it an elegant example of how proofs streamline understanding of calculus concepts. By breaking problems down into these logical steps, students gain confidence in their ability to handle more complex proofs in calculus.