How do you find the Limit of #x+3# as #x->2# and then use the epsilon delta definition to prove that the limit is L?

The epsilon delta proof for limits is easier understood when one is familiar with the definitions of the terms involved. Most useful will be the definition of the limit of a function.

Definition:

In our case:

So:

Then we know that we must have:

Note that all of the above steps come from the definition of the limit of a function as provided above.

To find the limit of x+3 as x approaches 2, we substitute the value of x into the expression x+3, which gives us 2+3=5.

To prove that the limit is L using the epsilon-delta definition, we need to show that for any given epsilon (ε) greater than 0, there exists a corresponding delta (δ) greater than 0 such that if 0 < |x-2| < δ, then |(x+3)-L| < ε.

Let's assume L = 5.

Now, we need to find a suitable delta that satisfies the condition.

Given ε > 0, we want to find δ > 0 such that if 0 < |x-2| < δ, then |(x+3)-5| < ε.

Simplifying the expression, we have |x-2| < δ implies |x-2| < ε.

Since we want to find a delta that satisfies this condition, we can choose δ = ε.

Now, if 0 < |x-2| < δ = ε, then |(x+3)-5| = |x-2| < ε.

Therefore, we have shown that for any given ε > 0, there exists a corresponding δ > 0 (in this case, δ = ε) such that if 0 < |x-2| < δ, then |(x+3)-5| < ε.

Hence, using the epsilon-delta definition, we have proven that the limit of x+3 as x approaches 2 is indeed 5.