# How do you prove that the limit of #(x^2) =9 # as x approaches -3 using the epsilon delta proof?

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To prove that the limit of (x^2) as x approaches -3 is 9 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if 0 < |x - (-3)| < delta, then |(x^2) - 9| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a suitable delta > 0.

We start by considering the expression |(x^2) - 9| and try to manipulate it to fit our requirements.

|(x^2) - 9| = |x + 3||x - 3|

Now, we can see that if |x + 3| < 1, then |(x^2) - 9| = |x + 3||x - 3| < (1)(|x - 3|).

So, we can choose delta = 1.

Now, let's assume that 0 < |x - (-3)| < delta = 1.

|x - (-3)| = |x + 3| < 1.

From this, we can deduce that -4 < x < -2.

Now, let's consider |(x^2) - 9| again:

|(x^2) - 9| = |x + 3||x - 3| < (1)(|x - 3|).

Since -4 < x < -2, we can conclude that -1 < x - 3 < 1.

Therefore, |(x^2) - 9| < (1)(|x - 3|) < 1.

Hence, we have shown that for any epsilon > 0, we can choose delta = 1, and if 0 < |x - (-3)| < delta, then |(x^2) - 9| < epsilon.

Therefore, the limit of (x^2) as x approaches -3 is indeed 9, as proven 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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