# How do you prove that the limit of #(2x^2 + 1) = 3 # as x approaches 1 using the epsilon delta proof?

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To prove that the limit of (2x^2 + 1) is 3 as x approaches 1 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 - 1| < delta, then |(2x^2 + 1) - 3| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a delta > 0 such that if 0 < |x - 1| < delta, then |(2x^2 + 1) - 3| < epsilon.

First, let's simplify the expression |(2x^2 + 1) - 3|:

|(2x^2 + 1) - 3| = |2x^2 - 2| = 2|x^2 - 1| = 2|x - 1||x + 1|

Now, we can set a condition for delta based on epsilon. Let's choose delta = min(1, epsilon/5).

Now, let's consider the case when 0 < |x - 1| < delta:

|x - 1| < delta |x - 1| < min(1, epsilon/5) |x - 1| < 1 (since delta is at most 1) |x - 1| < epsilon/5 (since delta is at most epsilon/5)

Now, let's analyze the expression |(2x^2 + 1) - 3|:

|(2x^2 + 1) - 3| = 2|x - 1||x + 1|

Since we have already established that |x - 1| < epsilon/5, we can substitute it into the expression:

2|x - 1||x + 1| < 2(epsilon/5)|x + 1|

Now, let's consider the case when |x + 1| < 3:

|x + 1| < 3 2(epsilon/5)|x + 1| < 2(epsilon/5)(3) 2(epsilon/5)|x + 1| < 6(epsilon/5) 2(epsilon/5)|x + 1| < 2epsilon

Therefore, if we choose delta = min(1, epsilon/5), we can conclude that if 0 < |x - 1| < delta, then |(2x^2 + 1) - 3| < epsilon.

Hence, we have proven that the limit of (2x^2 + 1) is 3 as x approaches 1 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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