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

Preliminary analysis

Look at the thing we want to make small. We want to see the thing we control.

Now we need to actually write up the proof:

Proof

We can condense a bit

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To prove that the limit of 5 - 2x = 1 as x approaches 2 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 if 0 < |x - 2| < delta, then |(5 - 2x) - 1| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a delta > 0 such that if 0 < |x - 2| < delta, then |(5 - 2x) - 1| < epsilon.

First, let's manipulate the expression |(5 - 2x) - 1| < epsilon:

|(5 - 2x) - 1| < epsilon |4 - 2x| < epsilon |-2(2 - x)| < epsilon 2|x - 2| < epsilon

Now, we can see that if we choose delta = epsilon/2, then:

0 < |x - 2| < delta 0 < |x - 2| < epsilon/2 0 < 2|x - 2| < epsilon

Therefore, we have shown that for any given epsilon > 0, there exists a corresponding delta > 0 (specifically, delta = epsilon/2) such that if 0 < |x - 2| < delta, then |(5 - 2x) - 1| < epsilon.

Hence, we have proven that the limit of 5 - 2x = 1 as x approaches 2 using the epsilon-delta proof.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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