# How do you prove that the limit of #(x^2 - x) = 0 # as x approaches 1 using the epsilon delta proof?

See the explanation section below.

Preliminary analysis

By definition,

Look at the thing we want to make small. Factor this, looking for the thing we control.

Now we need to actually write up the proof:

Proof

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To prove that the limit of (x^2 - x) as x approaches 1 is 0 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 - 1| < delta, then |(x^2 - x) - 0| < epsilon.

Let's proceed with the proof:

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

First, let's simplify |(x^2 - x) - 0| to |x^2 - x|.

Now, we can factor |x^2 - x| as |x(x - 1)|.

Since we are interested in values of x close to 1, we can assume that |x - 1| < 1, which implies -1 < x - 1 < 1.

Using the triangle inequality, we have |x(x - 1)| ≤ |x| |x - 1|.

Since we want to find a delta such that |(x^2 - x) - 0| < epsilon, we can set |x| |x - 1| < epsilon.

Now, we can choose a delta such that whenever 0 < |x - 1| < delta, we have |x| |x - 1| < epsilon.

Let's choose delta = min(1, epsilon).

If 0 < |x - 1| < delta, then we have -1 < x - 1 < 1, which implies -1 < x < 2.

Since we assumed |x - 1| < 1, we can also assume that 0 < x < 2.

Now, let's consider |x| |x - 1| < epsilon.

Since 0 < x < 2, we have |x| < 2.

Therefore, |x| |x - 1| < 2 |x - 1|.

Since we chose delta = min(1, epsilon), we have 2 |x - 1| < 2 delta ≤ 2 epsilon.

Hence, whenever 0 < |x - 1| < delta, we have |(x^2 - x) - 0| = |x| |x - 1| < 2 epsilon.

This completes the epsilon-delta proof, showing that the limit of (x^2 - x) as x approaches 1 is indeed 0.

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