# How do you prove that the limit of #((x^2+x-6)/(x-2))=5# as x approaches 2 using the epsilon delta proof?

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To prove that the limit of ((x^2+x-6)/(x-2)) equals 5 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 whenever 0 < |x - 2| < delta, then |((x^2+x-6)/(x-2)) - 5| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a delta > 0 such that |((x^2+x-6)/(x-2)) - 5| < epsilon whenever 0 < |x - 2| < delta.

First, let's simplify the expression ((x^2+x-6)/(x-2)) - 5:

((x^2+x-6)/(x-2)) - 5 = ((x^2+x-6) - 5(x-2))/(x-2) = (x^2+x-6 - 5x + 10)/(x-2) = (x^2 - 4x + 4)/(x-2) = (x-2)(x-2)/(x-2) = x-2.

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

Let's choose delta = epsilon.

If 0 < |x - 2| < delta = epsilon, then |x - 2| < epsilon.

Therefore, we have shown that for any given epsilon > 0, there exists a corresponding delta > 0 (specifically, delta = epsilon) such that whenever 0 < |x - 2| < delta, then |((x^2+x-6)/(x-2)) - 5| < epsilon.

Hence, we have proven that the limit of ((x^2+x-6)/(x-2)) equals 5 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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