EPSILON DELTA --PROOF -Calc Help?

Anonymous @Anonymous

May 2021 3 11 Report

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?

Science & Mathematics / Mathematics

Buri

Verified answer

Well in these epsilon-delta proofs your always trying to show that |f(x) - L| < ε. And so you "must" choose that |f(x) - L| < ε/2 and that |g(x) - M| < ε/2. Otherwise (if you had kept that they were both < ε), when you'd add them up as a consequence of the triangle inequality you'd be getting that |f(x) + g(x) - (L + M)| < 2ε which is not what the definition states (your trying to show < ε). So you must make it < ε/2. See when my professor did this proof, he went till the end with < ε and then ended up getting < 2ε so he told us that we should always make adjustments and go back to make the proof better. And he made the adjustments of making it < ε/2.

The min(δ1, δ2) that is to guarantee that they are BOTH true. And then leaves us only working with a single delta rather then two. This works because if lets say δ1 works for f(x) and δ2 is less then δ1 then for sure δ2 will work for f(x) also. Therefore, we can forget about δ1 and simply use δ2 for both of them. And that is exactly what the min(δ1, δ2) does.

Don't stress out with these epsilon-delta proofs. It's a very difficult concept which is very unlike anything in high school. It takes some time to master it well, so if you don't get it just yet leave it rest, and then come back to it. Try proving statements as well for practice.

Read the 5th chapter of this book, it should help a lot, it's actually the text I used for my analysis class:

http://books.google.com/books?id=7JKVu_9InRUC&prin...

Scranderberry showed you what your trying to show, but geometrically your trying to show that the summation of both of these functions also approaches the summation of their limits as x approaches a (i.e. you can make the summation of the two functions as close as you want to L + M by making x sufficiently close to a).

Hope this helps!

p.s. Always write things in terms of limit notation and then translate it into the precise definition of the limit. Make sure you understand what the definition means intuitively and remember always draw diagrams because this often times helps choosing the right delta.

garbarino

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