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

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To prove that the limit of ((x^2+x-6)/(x-2)) as x approaches 2 is equal to 5 using the formal definition of a limit, we need to show that for any given positive value ε, there exists a positive value δ such that if 0 < |x - 2| < δ, then |((x^2+x-6)/(x-2)) - 5| < ε.

Let's proceed with the proof:

Given ε > 0, we need to find a δ > 0 such that if 0 < |x - 2| < δ, then |((x^2+x-6)/(x-2)) - 5| < ε.

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 δ > 0 such that if 0 < |x - 2| < δ, then |x - 2| < ε.

Since we want to prove that the limit is 5, we can choose δ = ε.

If 0 < |x - 2| < δ = ε, then |x - 2| < ε.

Therefore, we have shown that for any given positive value ε, there exists a positive value δ (in this case, δ = ε) such that if 0 < |x - 2| < δ, then |((x^2+x-6)/(x-2)) - 5| < ε.

Hence, using the formal definition of a limit, we have proven that the limit of ((x^2+x-6)/(x-2)) as x approaches 2 is equal to 5.

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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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