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Analysis Proof Help :S

  1. Jan 25, 2004 #1
    Hi, im taking my first analysis course and we are studying Limits right now. My prof said they are the most important thing to remember out of this whole semester. Anyways i have two problems im trying to solve that i could do with some help.

    1) Show that if [tex]f(x) \leq 0[/tex] and [tex]\lim_{x->a} f(x) = l[/tex], then [tex]l \leq 0[/tex].

    2) If [tex]f(x) \leq g(x)[/tex] for all x, then [tex]\lim_{x->a} f(x) \leq \lim_{x->a} g(x)[/tex].
    If those limits exist.

    For number one i can see this is obvious but i don't know where to start to try and prove it.

    I know the definitions for limits, do i use them somehow?

    For number 1) i think that i can use a proof by contradiction somehow.
     
  2. jcsd
  3. Jan 25, 2004 #2

    Hurkyl

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    Yah, using the definition of limit will probably be helpful. Proof by contradiction sounds like a good plan, let us know how far you get!
     
  4. Jan 25, 2004 #3
    I have been working ont he second question:

    2) If [tex]f(x) \leq g(x)[/tex] for all x, then [tex]\lim_{x->a} f(x) \leq \lim_{x->a} g(x)[/tex].
    If those limits exist.

    Suppose that [tex]f(x) \leq g(x)[/tex]. Let [tex]h(x) = g(x)-f(x)[/tex]. So [tex]h(x) \geq 0[/tex].
    [tex]\lim_{x->a} h(x) = \lim_{x->a} g(x) - \lim_{x->a} f(x) \geq 0[/tex]. Therefore [tex]\lim_{x->a} f(x) \leq \lim_{x->a} g(x)[/tex]

    Is this right?

    Can someone give me a hint for question 1?
    Start me off or something......
     
    Last edited: Jan 25, 2004
  5. Jan 25, 2004 #4

    Hurkyl

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    What is your justification for this?



    Your hint for problem (1) is to assume [itex]l > 0[/itex] and rewrite the entire problem in terms of epsilons and deltas; exactly what you said you thought you should do. :wink:
     
  6. Jan 26, 2004 #5
    Ok i had a good nights sleep went to school and came back tonight with a fresh mind to give it another shot. And i believe it worked :D

    1) Show that if [tex]f(x) \leq 0[/tex] and [tex]\lim_{x->a} f(x) = l[/tex], then [tex]l \leq 0[/tex].

    [tex]\forall\epsilon > 0, \exists\delta > 0 \ni[/tex] for all [tex]x[/tex] , [tex]0< |x - a| < \delta[/tex], then [tex]|f(x) - l| < \epsilon[/tex].

    Assume that [tex]l > 0[/tex]

    [tex]\forall\epsilon > 0, \exists\delta > 0 \ni[/tex] for all [tex]x[/tex] , [tex]0< |x - a| < \delta[/tex], then [tex]l - f(x) < \epsilon[/tex].

    since [tex]l - f(x) > 0 \implies 0 < l - f(x) < \epsilon[/tex]

    add [tex]f(x)[/tex] to each side to get [tex]f(x) < l < \epsilon + f(x)[/tex].

    Choose [tex]\alpha[/tex] to be the distance from [tex]f(x)[/tex] to [tex]0[/tex]. Then we choose [tex]\delta \ni \epsilon = \alpha[/tex], then [tex]\epsilon + f(x) = 0[/tex]
    therefore, [tex]f(x) < l < 0[/tex] which is a contradiction to [tex]l > 0[/tex].

    Im sure this must be correct.

    I will try number 2 now. Thanks for your hint ;)
     
  7. Jan 26, 2004 #6

    Hurkyl

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    You're almost there: given the assumption [itex]l > 0[/itex], you've correctly proven

    [tex]\forall \epsilon > 0 \exists \delta > 0 \forall x: 0 < |x - a| < \delta \implies f(x) < l < f(x) + \epsilon[/tex]

    however, the next steps are wrong. For instance, the distance from [itex]f(x)[/itex] to zero could, in fact, be zero. (e.g. take [itex]f(x) = 0[/itex], or [itex]f(x) = -|x|[/itex])

    I'm also not sure what you're trying to do when you say "choose [itex]\delta[/itex] such that [itex]\epsilon = \alpha[/itex]"... when deriving the contradiction you do get to choose [itex]\epsilon[/itex], but the contradiction must hold for all [itex]\delta[/itex].

    You need another hint on this one?


    As for number two, the proof you gave almost works, now that you've proven #1... see if you can rewrite it to take advantage of knowing #1. :smile:
     
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