# How do you use the epsilon delta definition of limit to prove that #lim_(x->5)(x-1)= 4# ?

Let us review the definition of a limit first.

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To prove that lim_(x->5)(x-1)= 4 using the epsilon delta definition of limit, we need to show that for any given epsilon > 0, there exists a delta > 0 such that if 0 < |x-5| < delta, then |(x-1) - 4| < epsilon.

Let's proceed with the proof:

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

Consider |(x-1) - 4| = |x-5|.

We want to ensure that |x-5| < epsilon whenever 0 < |x-5| < delta.

Choosing delta = epsilon, we can see that if 0 < |x-5| < delta, then |x-5| < epsilon.

Hence, we have shown that for any epsilon > 0, there exists a delta > 0 such that if 0 < |x-5| < delta, then |(x-1) - 4| < epsilon.

Therefore, lim_(x->5)(x-1) = 4.

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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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