EPSILON DELTA --PROOF -Calc Help?

If lim x->a f(x) = L and lim x->a g(x)=M then lim x->a [f(x)+g(x)] = L + M

The book says:

Let E > 0 be given. We must find an D >0 such that

| f(x) +g(x)-(L+M) | < E whenever 0< | x-a| < D

It uses the triangle inequality and says

| f(x) +g(x)-(L+M) | <=|f(x)-L| + | g(x)-M|

Now the parts I dont get

It makes | f(x) +g(x)-(L+M) | < E by making |f(x)-L| and | g(x)-M| < E/2

and then E/2 > 0 then it rewrites the def for a D1 and D2 for

| f(x)-L | <E/2 whenever 0 <|x-a|<D1 and |g(x)-M|<E/2 whenever 0<|x-a|<E/2 (I dont really get why the chose E/2)

then I really really dont understand this part

It lets D = min{D1,D2} ---> No idea --I dont get this part at alll

if 0<|x-a| <D then 0 <|x-a|<D1 and0<|x-a|<D2

and so |f(x)-L|<E/2 and |g(x)-M|<E/2

I really dont get that part ...can someone help me please?

Thanks

Update:

I just looked at it again and its starting to make more sense

I just dont get why it does the min thing and how do I know to choose E/2

..what am I really trying to show?

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