We actualy want to prove:

For that you must prove: given ε>0 there exists a δ>0 such that:

..........if xεA and 0<|x-c|<δ , then . The trick of that case is hiding in the last inequality:

........................... .......................1

Ιf we now multiply by and at the same time divide by we get:

.

Now since f(x)>0 ====> >0=====> .

Hence < .......................======>

< .

Overall now we have :

.................................................. 2

This last inequality is very important to the proof that follows:

Note the work we have done so far can be considered as an aside work.

Now for the proof :

Let ε>0 then .

But , since ,then for any E>0 we can find a δ>0 such that :

........................if xεA and 0<|x-c|<δ ,then |f(x)-m|<E...............

Since the above holds for any E>0,it will hold for .

And for that E we can find a δ>0, such that:

...........if xεA and 0<|x-c|< δ,then |f(x)-m|< .................................................. .3

Now we are nearly done,so let:

...........xεA and 0<|x-c|<δ.............................................

............then in combination of (2) and (3) we get the desired result:

................................ .................................................. .....................

Of course if one can write a more detailed proof the more he/she gets to the inside of the proof,and i think this should be the ultimate goal of every student in analysis,otherwise one may run into the possibility of ,what you may remember one year you may have forgotten the next.