# How do you prove that the limit of #1/(x-3)# does not exist as x approaches 3 using the epsilon delta proof?

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To prove that the limit of 1/(x-3) does not exist as x approaches 3 using the epsilon-delta proof, we need to show that for any given epsilon value, there does not exist a corresponding delta value that satisfies the definition of a limit.

Let's assume that the limit L exists. According to the epsilon-delta definition, for any epsilon greater than 0, there exists a delta greater than 0 such that if 0 < |x - 3| < delta, then |1/(x-3) - L| < epsilon.

To disprove the existence of the limit, we need to find an epsilon value for which there is no corresponding delta value. Let's consider the case when epsilon is equal to 1.

Now, we need to show that for any delta value chosen, there exists an x value such that 0 < |x - 3| < delta, but |1/(x-3) - L| ≥ 1.

Let's assume L is a real number. If we choose x = 3 + delta/2, then |x - 3| = delta/2 < delta. However, |1/(x-3) - L| = |1/(delta/2) - L| = |2/delta - L| ≥ 1, which contradicts the definition of the limit.

Therefore, we have shown that for epsilon = 1, there does not exist a corresponding delta value that satisfies the definition of a limit. Hence, the limit of 1/(x-3) does not exist as x approaches 3 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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