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Need help Proving infinite limit properties.

  1. Sep 5, 2011 #1
    Suppose that limx->c f(x) = infinity and limx->c g(x)=l where l is a real number. Prove the following.
    limx->c[f(x)+g(x)]= infinity
    limx->c[f(x)g(x)]= infinity if l > 0
    limx->c[f(x)g(x)]= -infinity if l < 0

    I have the proof for these already, but I couldn't understand them, would someone please explain them.
    The thing I don't understand is where does the L come from, but explanation in general would be greatly appreciated

    Here is the proof for the second one:

    [URL=http://imageshack.us/photo/my-images/190/unledmkd.png/][PLAIN]http://img190.imageshack.us/img190/2665/unledmkd.png[/URL]

    Uploaded with ImageShack.us[/PLAIN]
     
    Last edited: Sep 5, 2011
  2. jcsd
  3. Sep 5, 2011 #2

    mathman

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    There seems to be a change of notation. In the original, g -> c when x -> a. In the equations in question, g -> L when x -> c.
     
  4. Sep 5, 2011 #3
    sorry, i corrected it, does anyone have the explanation for this proof ?
     
  5. Sep 5, 2011 #4
    What do you mean "where does the L come from"? It is the limit of the function g(x).

    I remind you, what you need to prove, is that given an M>0, you can find a [itex]\delta>0[/itex], so that for every x that satisfied |x-c|<[itex]\delta[/itex],
    |f(x)g(x)| > M hold.

    The proof shows the existence of such a [itex]\delta>0[/itex].

    In order to do this alone, however, you should start from the end. You should ask yourself, how can I make |f(x)g(x)| > M happen?
    You need to remember that f(x) can be as big as you want when you're close enough to c, and that g(x) can be as close to L as you'd like when you're close enough to c.

    That's exactly what they used in the proof.
     
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