Question

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

Answer 1

Before writing a proof, I would do some scratch work in order to find the expression for `delta` in terms of `epsilon`.

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

For all `epsilon > 0`, there exists `delta > 0` such that
`0<|x-1|< delta Rightarrow |(x+2)-3| < epsilon`.

Start with the conclusion.

`|(x+2)-3| < epsilon Leftrightarrow |x-1| < epsilon`

So, it seems that we can set `delta =epsilon`.

(Note: The above observation is just for finding the expression for `delta`, so you do not have to include it as a part of the proof.)

Here is the actual proof:

Proof

For all `epsilon > 0`, there exists `delta=epsilon > 0` such that
#0<|x-1| < delta Rightarrow |x-1|< epsilon Rightarrow |(x+2)-3| < epsilon#