Using the definition of a limit for real functions, prove thatlimx→2 x2 = 4

Sure, here is the proof using the definition of a limit for real functions:

The definition of a limit states that for every ε > 0, there exists a δ > 0 such that if 0 < |x - 2| < δ, then |x² - 4| < ε.

We want to show that the limit as x approaches 2 of x² is 4. That is, we want to find a δ such that if 0 < |x - 2| < δ, then |x² - 4| < ε.

Let's start by simplifying |x² - 4|. This simplifies to |(x - 2)(x + 2)|.

We know that if 0 < |x - 2| < δ, then -δ < x - 2 < δ. Adding 2 to all parts of the inequality gives 2 - δ < x < 2 + δ.

We can then say that x + 2 < 2 + δ + 2 = 4 + δ, so |x + 2| < 4 + δ.

Therefore, |(x - 2)(x + 2)| = |x - 2||x + 2| < δ(4 + δ).

We want this to be less than ε, so we need δ(4 + δ) < ε.

If we choose δ to be min(1, ε/5), then δ ≤ 1 and δ ≤ ε/5.

If δ ≤ 1, then 4 + δ ≤ 5, and if δ ≤ ε/5, then δ(5) ≤ ε.

Therefore, if 0 < |x - 2| < δ, then |x² - 4| < ε, which proves that the limit as x approaches 2 of x² is 4.

Sure, here is the proof using the definition of a limit for real functions:

The definition of a limit states that for every ε > 0, there exists a δ > 0 such that if 0 < |x - 2| < δ, then |x² - 4| < ε.

We want to show that the limit as x approaches 2 of x² is 4. That is, we want to find a δ such that if 0 < |x - 2| < δ, then |x² - 4| < ε.

Let's start by simplifying |x² - 4|. We can factor this as |(x - 2)(x + 2)|.

Now, we want to find a δ such that if 0 < |x - 2| < δ, then |(x - 2)(x + 2)| < ε.

We can make this easier by bounding |x + 2|. Since we're taking the limit as x approaches 2, we can say that x is near 2, so |x + 2| is near 4. Therefore, we can say that |x + 2| < 5 for x near 2.

So, if 0 < |x - 2| < δ, then |(x - 2)(x + 2)| = |x - 2| * |x + 2| < δ * 5.

We want this to be less than ε, so we can choose δ = ε/5. Then, if 0 < |x - 2| < δ, we have |x² - 4| = |(x - 2)(x + 2)| < δ * 5 = ε.

Therefore, by the definition of a limit, the limit as x approaches 2 of x² is 4.