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How to prove this?

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello! Now I want to prove this using only δ and ε:[tex]\lim_{x\rightarrow p}f(x)+g(x)=\lim_{x\rightarrow p}f(x)+\lim_{x\rightarrow p}g(x)[/tex]


    2. Relevant equations



    3. The attempt at a solution
    These are my attempts to solve the problem.
    First we start with
    [tex]\lim_{x\rightarrow p}f(x)+g(x)[/tex]
    if the limit do exist (and we know it does) then
    [tex]p-\delta <x<p+\delta \Rightarrow f(p)-\epsilon <f(x)+g(x)<f(p)+\epsilon[/tex]
    and it's limit is given by
    [tex]0<x-p<\delta \Rightarrow f(x)+g(x)-L<\epsilon[/tex]
    [tex]0<x-p<\delta \Rightarrow L>-\epsilon+f(x)+g(x)[/tex]

    now we do the same thing for f(x)
    if the limit do exit then
    [tex]p-\delta <x<p+\delta \Rightarrow f(p)-\epsilon <f(x)<f(p)+\epsilon[/tex]
    and it's limit is given by
    [tex]0<x-p<\delta \Rightarrow f(x)-L<\epsilon[/tex]
    [tex]0<x-p<\delta \Rightarrow L>-\epsilon+f(x)[/tex]

    now we do it for g(x)
    [tex]p-\delta <x<p+\delta \Rightarrow g(p)-\epsilon <g(x)<g(p)+\epsilon[/tex]
    and it's limit is given by
    [tex]0<x-p<\delta \Rightarrow g(x)-L<\epsilon[/tex]
    [tex]0<x-p<\delta \Rightarrow L>-\epsilon+g(x)[/tex]

    Now we see that [tex]L>-\epsilon+g(x)[/tex] and [tex]L>-\epsilon+f(x)[/tex] hence
    [tex]f(x)=g(x)[/tex]
    Now I'm stuck! I mean, I don't know if what I did is correct so I don't think I should go on...
     
    Last edited: Apr 23, 2012
  2. jcsd
  3. Apr 23, 2012 #2
    I think they want you to prove that

    [tex]\lim_{x\rightarrow p}(f(x)+g(x))=\lim_{x\rightarrow p}f(x)+\lim_{x\rightarrow p}g(x)[/tex]

    Big difference.
     
  4. Apr 23, 2012 #3
    Yes! Sorry my fault.
    Anyway, do you know how to prove it?
     
  5. Apr 23, 2012 #4

    SammyS

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    I assume that what need to do is the following.
    If [itex]\displaystyle \lim_{x\to \ p}f(x)[/itex] and [itex]\displaystyle \lim_{x\to\ p}g(x)[/itex] exist, then prove that [itex]\displaystyle \lim_{x\to\ p}(f(x)+g(x))=\lim_{x\to\ p}f(x)+\lim_{x\to\ p}g(x)\ .[/itex]​
    Given ε > 0, use εf = ε/2 and εg = ε/2 .

    You then know that there is some δf and δg .

    What should your δ be for the limit of the sum ?
     
  6. Apr 23, 2012 #5
    Why εf = ε/2 and εg = ε/2?

    My δ would be δ=min{δf,δg}, correct?
     
  7. Apr 23, 2012 #6
    Yes. I already took that class!
     
  8. Apr 23, 2012 #7

    SammyS

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    That's the correct δ.

    Put it all together to see why εf = ε/2 and εg = ε/2 .
     
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