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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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.
the biggest element you'll run into once you pass to college is the sheer quantity of work. this is critically more effective than what you're used to. for brand spanking new recommendations, you'll also probably could answer questions about maximizing or minimizing (excuse me - maximising or minimising) a fee in a note difficulty. also looking the quantity of a sturdy created via rotating a function around the x or y axis. as an social gathering: At $a million.00 in accordance to glass, a student will promote seventy 5 glasses of lemonade in accordance to day. for each $0.10 she will boost the fee, she'll promote 5 fewer glasses. What fee will maximize sales? locate the quantity of the sturdy created even as rotating y = x^2 around the x-axis between x=a million and x=3.
you do the min thing because you want it to be true for both
And you use E/2 so that they dont overlap
what you are trying to show is this:
If lim x->a f(x) = L and lim x->a g(x)=M then lim x->a [f(x)+g(x)] = L + M