Question

Prove the given limit using an \(\varepsilon-\delta\) proof. $$ \lim _{x \rightarrow 2} 5=5 $$

Short answer

Constant function always satisfies the limit; choose any \( \delta > 0 \).

Step-by-step solution

Understand the Limit Definition

The definition of a limit states that \( \lim_{x \to a} f(x) = L \) if for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x-a| < \delta \), it follows that \( |f(x) - L| < \varepsilon \). Here, \( a = 2 \), \( L = 5 \), and \( f(x) = 5 \).

Analyze the Function

The function \( f(x) = 5 \) is a constant function. Therefore, irrespective of the value of \( x \), \( f(x) = 5 \). Thus, \( |f(x) - 5| = |5 - 5| = 0 \) for any \( x \).

Choose Delta

Since \( |f(x) - 5| = 0 \) is always true, we can choose any positive \( \delta \). Typically, in this situation, you can simply choose \( \delta = 1 \) or any other positive number, because \( |f(x) - 5| < \varepsilon \) is trivially satisfied for any \( \varepsilon > 0 \).

Apply the Epsilon-Delta Definition

Given any \( \varepsilon > 0 \), since \( |f(x) - 5| = 0 < \varepsilon \), we confirm that for any \( 0 < |x - 2| < \delta \), the condition \( |f(x) - 5| < \varepsilon \) holds true. This completes the \( \varepsilon-\delta \) proof.

Key concepts

Limit of a Function

Understanding the concept of the "limit of a function" is fundamental in calculus. Limits help us grasp the behavior of a function as the input approaches a certain value. This concept is essential for defining other concepts in calculus such as continuity, derivatives, and integrals.
Think of a limit as aiming to understand what the value of a function approaches as the input gets infinitely close to a specific point. A formal definition would say that \( \lim_{x \to a} f(x) = L \) if, for every arbitrarily small number \( \varepsilon > 0 \), there is a corresponding distance \( \delta > 0 \) where if \( x \) is within \( \delta \) of \( a \) \( \text{(excluding the actual point} \; a\text{)} \), then the value of \( f(x) \) will be within \( \varepsilon \) of \( L \).

• \( \varepsilon \) represents how close \( f(x) \) needs to be to \( L \).
• \( \delta \) represents how close \( x \) is to \( a \).

For the limit \( \lim_{x \rightarrow 2} 5 = 5 \), this means as \( x \) approaches 2, the function remains constant at 5. This illustrates how limits help us concretely define the behavior of functions near particular points.

Constant Function

A constant function is a special type of function where the output value is the same no matter the input. Mathematically, if \( f(x) = c \), then no matter the value of \( x \), \( f(x) = c \). This implies that the function is a horizontal line on a graph.
For this specific example, \( f(x) = 5 \), so no matter what value \( x \) takes, \( f(x) \) always equals 5. Therefore, the behavior of the function doesn't change when \( x \) approaches any number, including 2. This makes the limit definition straightforward since the \( \varepsilon-\delta \) inequality \( |f(x) - 5| < \varepsilon \) is always satisfied, regardless of the values of \( \varepsilon \) or \( \delta \).

• The "closeness" requirement in limits is trivially met because the function does not vary.
• Typically, constant functions have very simple limits because their value does not depend on \( x \).

Constant functions, by their nature, simplify many limit proofs because they eliminate variability, thus bypassing the complexities involved in analyzing changing functions.

Mathematical Proof

Mathematical proofs provide clarity and verification, demonstrating that statements or propositions are true beyond any doubt through a logical series of steps. An epsilon-delta proof is a classic method in calculus used to rigorously show the limit of a function. It's an essential proof technique because it uses precise definitions and logic to confirm conclusions.
To work through an epsilon-delta proof, here's a streamlined approach:

• Start by defining your target limit in terms of \( \varepsilon \) and \( \delta \).
• Show that for any given \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x-a| < \delta \), \( |f(x) - L| < \varepsilon \).
• Conclude by confirming the condition holds true for all values within the defined range.

In the example limit \( \lim_{x \rightarrow 2} 5 = 5 \), because the function is constant, the proof is quite direct. Since \( |f(x) - 5| = 0 \) for all values of \( x \), any chosen \( \delta \) works. This illustrates how mathematical proofs not only affirm results but simplify understanding by laying out every detail explicitly.