About the definition of the limit, it is a very very natural definition. As I have already said a limit exists if you can show me that you can get as close as you want to it, Read want I wrote in my earlier posts and compare it to the definition, (maybe it might help for you to draw a diagram of the regions represented to make it more clear).

Now for this example given here, what you are proving is slightly silly because the function under discussion is continuous (maybe you have not got onto this chapter yet, but by definition the limit of a continuous function at a point and its value are the same). This is how your present the argument for this question.

We are given

and we want to find a

such that when we are in the region

we have

.

What we notice is that when we have x such that

then

(this should be obvious, if it is not draw out the regions of do the algebra),

so we get

Happy so far ? good,

now we want

Remember we are working backwards here, we want to show that you any epsilon we can find delta i.e. give any epsilon we can find delta, so we need delta in terms of epsilon. so at this point you have two choice, you can either solve the quadratic equation

, or do what the author of your book did and treat the cases

and

separately in order to avoid solving the equation.

The only problem I see is that in your text the authord neglected the case

, This cases is somewhat trivial though.

I am surprised you are asking this question, because any decent text on analysis will discuss this at some point. I'll briefly tell you how you deal with this.

now we apply the triangle inequality

I want for any epsilon to have

and

that is fine, because we know those limits exist so we get

and

such that:

and now we pick

giving

so we can make the difference between L and M as small as we like.