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

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To prove that the limit of (2-x)/(1+x^2) is 3/2 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 |(2-x)/(1+x^2) - 3/2| < epsilon.

Let's proceed with the proof:

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

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

|(2-x)/(1+x^2) - 3/2| = |(4-2x-x^3)/(2+2x^2)|

Now, we can set an upper bound for the numerator:

|4-2x-x^3| ≤ 4 + 2|x| + |x^3|

Since we are interested in values of x close to 1, we can assume that |x| < 1. Therefore, we have:

|4-2x-x^3| ≤ 4 + 2|x| + |x^3| ≤ 4 + 2 + 1 = 7

Now, let's set a lower bound for the denominator:

2+2x^2 > 2

Combining the upper bound for the numerator and the lower bound for the denominator, we have:

|(2-x)/(1+x^2) - 3/2| ≤ (7/2) / 2 = 7/4

Now, we can choose delta as the minimum between 1 and (4/7) * epsilon:

delta = min(1, (4/7) * epsilon)

By choosing this delta, we can ensure that if 0 < |x - 1| < delta, then |(2-x)/(1+x^2) - 3/2| < epsilon.

Therefore, we have proven that the limit of (2-x)/(1+x^2) is 3/2 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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