We need to show that given any `epsilon>0,` we can find some `delta>0` such that `|(3x+5)-(-1)|=|3x+6|<epsilon` whenever `0<|x-(-2)|=|x+2|<delta.`
As usual in this type of proof, we essentially work backwards to get a value of `delta` and then reverse our steps to come up with an acceptable proof.
`|3x+6|=|3(x+2)|=3|x+2|<epsilon` implies `|x+2|<epsilon/3,` which means that given some value `epsilon,` we should try choosing `delta=epsilon/3.` Then we reverse the above steps to come up with the following proof:
Let `epsilon>0` be given. Set `delta=epsilon/3.` Then if `0<|x+2|<epsilon/3,` we have
`epsilon>3|x+2|=|3x+6|=|(3x+5)-(-1)|,` which is what we needed to show.