Step 1: Define the Limit Statement
In calculus, when we talk about the limit of a function at a certain point, we’re essentially asking what value the function approaches as its input (or ‘x’ value) gets closer and closer to that point. In this specific case, we’re interested in the function f(x) = x² and we want to find out what it approaches as x approaches 2.
The formal way to express this is to say that we want to prove that the limit of f(x) = x² as x approaches 2 is 4. This is often written as lim (x -> 2) x² = 4.
But what does it mean for a limit to ‘be’ a certain number? It means that as x gets arbitrarily close to 2, f(x) will get arbitrarily close to 4. We need to show that no matter how small a ‘gap’ ε we specify around 4, there’s a corresponding ‘gap’ δ around 2 such that if x is within that δ-gap around 2 (but not exactly at 2), then f(x) will be within the ε-gap around 4.
In even more formal terms, we want to show that for every ε > 0, there exists a δ > 0 such that if 0 < |x - 2| < δ, then |x² - 4| < ε. This is the precise mathematical statement we aim to prove.
Step 2: Rewrite the Function Expression
Our goal is to find a relationship between ε (epsilon) and δ (delta) that makes our limit statement true. To do this, we need to work with the expression |x² – 4|, as this is what we’re trying to bound by ε.
First, let’s look at the expression |x² – 4|. Why is this expression important? Because it represents the difference between the function value f(x) = x² and the limit value 4. We want this difference to be less than ε.
Now, we rewrite |x² – 4| as |(x – 2)(x + 2)|. Why do we do this? Because it introduces the term |x – 2|, which is what we want to relate to ε. This is a common algebraic technique to simplify complex expressions. It’s like breaking down a complex machine into its basic parts to understand how it works. In this case, we’re breaking down the expression |x² – 4| to understand how it relates to |x – 2|.
So, we have |x² – 4| = |(x – 2)(x + 2)|. This is our starting point for finding the relationship between ε and δ.
Step 3: Establish the Goal for ε and δ
Now that we have rewritten |x² – 4| as |(x – 2)(x + 2)|, our next task is to find a relationship between ε (epsilon) and δ (delta) that makes our limit statement true. Remember, our ultimate goal is to show that for every ε > 0, there exists a δ > 0 such that if 0 < |x - 2| < δ, then |x² - 4| < ε.
Why is this our goal? Because this is the formal definition of a limit. We need to show that we can make |x² – 4| as close to 0 as we want (i.e., less than any given ε) by making |x – 2| sufficiently small (i.e., less than some δ).
Our task now is to find a relationship between δ and ε that makes this statement true. We start with the inequality |(x – 2)(x + 2)| < ε because this is what we want to prove. We look for ways to rewrite this inequality in terms of |x - 2|.
Why do we focus on the inequality |(x – 2)(x + 2)| < ε? Because it's the mathematical representation of our goal. It's like a puzzle that we need to solve, and solving it will prove that the limit exists and is equal to 4.
So, our starting point for this step is the inequality |(x – 2)(x + 2)| < ε. We will manipulate this inequality to find a suitable δ that works for our given ε.
Clarification on the +4 Term
Our goal is to find a relationship between ε and δ that makes our initial limit statement true. We started with the inequality \( |x^2 – 4| < \epsilon \) and rewrote it as \( |(x - 2)(x + 2)| < \epsilon \).
We rewrote \( |x^2 – 4| \) as \( |(x – 2)(x + 2)| \) to introduce \( x – 2 \) into the equation. This is important because we want to relate \( \epsilon \) and \( \delta \) through \( |x – 2| \).
Now, let’s focus on \( |x – 2| \). We know that \( |x – 2| < \delta \). This is our starting point. We want to express \( |x + 2| \) in terms of \( \delta \) as well.
Since \( |x – 2| < \delta \), we can rewrite this as \( -\delta < x - 2 < \delta \). Adding 2 to all sides of this inequality gives us \( -\delta + 2 < x < \delta + 2 \).
Now, let’s find an upper bound for \( |x + 2| \). Since \( x < \delta + 2 \), adding 2 to both sides gives \( x + 2 < \delta + 4 \). Therefore, \( |x + 2| < \delta + 4 \).
This is a crucial step because now we have both \( |x – 2| \) and \( |x + 2| \) expressed in terms of \( \delta \).
Finally, we rewrite \( |(x – 2)(x + 2)| \) as \( |x – 2| \times |x + 2| \). Since \( |x – 2| < \delta \) and \( |x + 2| < \delta + 4 \), we can say \( |x - 2| \times |x + 2| < \delta (\delta + 4) \).
To make \( |x^2 – 4| < \epsilon \) true, we can choose \( \delta \) such that \( \delta (\delta + 4) = \epsilon \). This completes our proof.