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Calculus, Analytical demonstration

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Demonstrate Analytically:
    If [itex]\lim_{x\to a}f(x)=L[/itex] and L<0, there exists a δ>0 so that x[itex]\in[/itex]Dom(f), 0<|x-a|<δ [itex]\Rightarrow[/itex] f(x)<0

    2. The attempt at a solution
    I start identifying what I want to demonstrate and what are my assumptions.
    x[itex]\in[/itex]Dom(f), 0<|x-a|<δ [itex]\Rightarrow[/itex] f(x)<0
    I'm aware I have to reach f(x)<0 from the fact that L<0 and using the definition of a limit, the x belonging to the domain is there so that it also covers the cases were if not stated, one could get a limit that does not exist due.

    So I have:
    0<|x-a|<δ and 0<|f(x)-L|<ε
    and I suspect the way is not in the delta part but working with 0<|f(x)-L|<ε, but because epsilon is in relation to delta it never banish, meaning I would have to work with the other part.

    0<|f(x)-L|
    Because L<0 is part of the hypothesis
    L<|f(x)-L|+L
    L-|f(x)-L|<L
    Triangle Inequality
    L-|f(x)|-|L|<L
    Because L<0 is part of the hypothesis
    L-|f(x)|-(-L)<L
    L-|f(x)|+L<L
    L-|f(x)|<0
    -|f(x)|<-L
    L<-f(x)<-L
    -L>f(x)>L
    And I don't know where else to go to get the f(x)<0 form, I simply have no way to get rid of the L in the process without being left with an absolute value on f(x).
     
  2. jcsd
  3. Apr 15, 2012 #2

    HallsofIvy

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    Using the defintion of limit, take [itex]\epsilon[/itex] to be any number less than L.
     
  4. Apr 15, 2012 #3
    It works wonderfully and makes it two simple steps. Thanks.

    Though I'm inclined to ask; is this something that can always be done, correlating epsilon with the actual limit?
     
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