# How do you prove that the limit of #(x^2 - 4x + 5) = 1# as x approaches 2 using the epsilon delta proof?

See below.

Proof:

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To prove that the limit of (x^2 - 4x + 5) as x approaches 2 is equal to 1 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a corresponding delta greater than 0 such that whenever 0 < |x - 2| < delta, then |(x^2 - 4x + 5) - 1| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a delta > 0 such that |(x^2 - 4x + 5) - 1| < epsilon whenever 0 < |x - 2| < delta.

First, let's simplify the expression |(x^2 - 4x + 5) - 1|: |(x^2 - 4x + 4)| = |(x - 2)(x - 2)| = |x - 2|^2

Now, we want to find a delta such that |x - 2|^2 < epsilon whenever 0 < |x - 2| < delta.

Taking the square root of both sides, we have: |x - 2| < sqrt(epsilon)

So, if we choose delta = sqrt(epsilon), then whenever 0 < |x - 2| < delta, we have |(x^2 - 4x + 5) - 1| < epsilon.

Therefore, we have proven that the limit of (x^2 - 4x + 5) as x approaches 2 is equal to 1 using the epsilon-delta proof.

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