# How do you use the epsilon delta definition of limit to prove that #lim_(x->1)(x+2)= 3# ?

According to the epsilon delta definition, we want to say:

Start with the conclusion.

Here is the actual proof:

Proof

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To prove that lim_(x->1)(x+2)= 3 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 - 1| < delta, then |(x+2) - 3| < epsilon.

Let's proceed with the proof:

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

Now, |(x+2) - 3| = |x - 1|.

We want to ensure that |x - 1| < delta implies |(x+2) - 3| < epsilon.

Let's choose delta = epsilon.

If 0 < |x - 1| < delta = epsilon, then |(x+2) - 3| = |x - 1| < epsilon.

Hence, we have shown that for any epsilon > 0, there exists a delta > 0 (specifically, delta = epsilon) such that if 0 < |x - 1| < delta, then |(x+2) - 3| < epsilon.

Therefore, by the epsilon delta definition of limit, we can conclude that lim_(x->1)(x+2) = 3.

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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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