Epsilon-Delta Proof: Limit Of 2x-1 As X→0

Nov 3, 2025 by Admin 42 views

Let's dive into the fascinating world of calculus and tackle a classic problem using the rigorous epsilon-delta definition of a limit. This might sound intimidating, but don't worry, we'll break it down step by step. We aim to prove that the limit of the function f(x) = 2x - 1 as x approaches 0 is indeed -1. This proof method is fundamental in understanding the core concepts of limits and continuity in calculus.

Understanding the Epsilon-Delta Definition

Before we jump into the proof, let's make sure we're all on the same page about what the epsilon-delta definition actually means. At its heart, it's a way to formalize the intuitive idea of a limit: as x gets closer and closer to a certain value (in our case, 0), the function's value f(x) gets closer and closer to another value (in our case, -1). Epsilon (ε) represents how close we want f(x) to be to the limit, and delta (δ) represents how close x needs to be to its target value to achieve that closeness. To really grasp this concept, think of it as a game. Someone gives you an epsilon, and your job is to find a delta that makes the definition work.

The formal definition states: For any ε > 0, there exists a δ > 0 such that if 0 < |x - a| < δ, then |f(x) - L| < ε. Here,

ε (epsilon) represents an arbitrarily small positive number. It defines the tolerance within which the function's output must lie.
δ (delta) represents another positive number, which depends on ε. It defines a neighborhood around the point a such that if x is within this neighborhood, then f(x) is within the ε-neighborhood of the limit L.
a is the value that x approaches.
L is the limit of the function as x approaches a.
f(x) is the function for which we are evaluating the limit.

Basically, what the epsilon-delta definition is saying, in plain English, is this: no matter how small a margin of error (epsilon) someone gives you around the limit, you can always find a distance (delta) around the input value that guarantees the function's output stays within that margin of error. It's all about controlling the function's output by controlling its input. The crux of the matter lies in finding the right δ for any given ε. This might seem abstract, but it provides a rock-solid foundation for calculus, allowing us to deal with infinitely small quantities rigorously. Visualizing this on a graph can be incredibly helpful. Think of a band of width 2ε centered around the limit value L on the y-axis. The goal is to find a corresponding interval of width 2δ centered around 'a' on the x-axis, such that the function's graph within this interval stays entirely within the epsilon band. This encapsulates the core idea of the epsilon-delta definition, making it a vital tool for defining and proving limits.

Setting Up the Proof for lim (x→0) (2x - 1) = -1

Now, let's apply this definition to our specific problem. We want to prove that: lim (x→0) (2x - 1) = -1. In our case, f(x) = 2x - 1, a = 0, and L = -1. This means we need to show that for any ε > 0, we can find a δ > 0 such that if 0 < |x - 0| < δ, then |(2x - 1) - (-1)| < ε. This sets the stage for our proof, clearly outlining what we need to demonstrate. The goal is to manipulate the inequality |(2x - 1) - (-1)| < ε to find a relationship between ε and |x - 0|, which will allow us to determine a suitable value for δ. This value of δ will depend on ε and ensure that whenever x is within δ units of 0, the function value 2x - 1 is within ε units of -1. Think of it like setting up a puzzle – we have all the pieces, and now we need to fit them together to reveal the solution. The key is to start with the epsilon inequality and work backward until we can express δ in terms of ε. This process involves algebraic manipulation and a bit of clever thinking, but the underlying logic is always the same: controlling the input to control the output. Successfully setting up the proof is half the battle, as it provides a clear roadmap for the subsequent steps.

The Epsilon-Delta Proof: Step-by-Step

Alright, let's get down to the nitty-gritty of the proof. Remember, our goal is to find a δ that works for any given ε. Here's how we do it:

Start with the epsilon inequality: We want to show that |(2x - 1) - (-1)| < ε. This is the heart of the matter, the condition we need to satisfy. The left-hand side represents the distance between the function's value and the limit, and we want this distance to be less than epsilon. Essentially, we're saying we want the function's output to be within a certain range of the limit.
Simplify the expression: |(2x - 1) - (-1)| simplifies to |2x|. This simplification is crucial because it connects the epsilon inequality to the expression involving x, which we can then relate to delta. By simplifying, we're making the problem more manageable and highlighting the key relationship between x and the distance from the limit.
Rewrite the inequality: Now we have |2x| < ε. This is equivalent to 2|x| < ε. Remember, we're trying to find a connection between |x - 0| (which is just |x|) and ε. This inequality brings us closer to that goal.
Isolate |x|: Divide both sides by 2 to get |x| < ε/2. This is a major breakthrough! We've now expressed a bound on |x| in terms of ε. This is exactly what we needed to find a suitable value for δ.
Choose δ: We can now choose δ = ε/2. This is the magic step! By setting delta to epsilon divided by 2, we ensure that whenever |x| is less than delta, the original epsilon inequality holds. In other words, we've found a way to control the function's output by controlling its input.
Formalize the proof: Now, let's write out the formal proof. Let ε > 0 be given. Choose δ = ε/2. If 0 < |x - 0| < δ, then |(2x - 1) - (-1)| = |2x| = 2|x| < 2(ε/2) = ε. Therefore, the limit as x approaches 0 of 2x - 1 is -1. This final step neatly ties everything together, presenting a clear and concise argument that satisfies the epsilon-delta definition.

Why This Proof Matters

Okay, you might be thinking,