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Math Help - Limits of Functions in Real Analysis

  1. #1
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    Limits of Functions in Real Analysis

    Alright, I need some guidance with the following problem. Any help would be appreciated.
    Suppose lim_{x\rightarrow x_{0}}f(x)=b and given any r>0 the set f(B'_{r}(a)) contains both poisitive and negative real numbers,show that b=0.

    After I talked to some people it sounds like there may be a typo???
    Possibly x_{0}=a in this problem? Either way I'm not really sure how to show that b=0.
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  2. #2
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    Quote Originally Posted by zebra2147 View Post
    Alright, I need some guidance with the following problem. Any help would be appreciated.
    Suppose lim_{x\rightarrow x_{0}}f(x)=b and given any r>0 the set f(B'_{r}(a)) contains both poisitive and negative real numbers,show that b=0.

    After I talked to some people it sounds like there may be a typo???
    Possibly x_{0}=a in this problem? Either way I'm not really sure how to show that b=0.

    It must be, again, a=x_0 : just think that when x\rightarrow x_0 , x gets into ANY neighborhood of x_0, so if in ANY such

    neighborhood there are positive and negative values of f AND the limit exists, this limit must be zero.

    Try now to give a proof with \epsilon\,,\,\delta and/or a proof bya contradiction.

    Tonio
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  3. #3
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    Attempt #1 at proof:

    Suppose D=D_{1}\bigcup D_{2} where D_{1} contains only negative real numbers and D_{2} contains only positive real values. Let a be a limit point of D_{1} and D_{2}. Then let f: D\rightarrow R be the function f(B'_{r}(a)). Thus, the lim_{x\longrightarrow a} exists only if lim_{x\longrightarrow a}f|_D_{j} both exist and are equal.

    Then, if both limits exist we have that, for some \epsilon>0,when x_{1}<0, |x_{1}-a|<\delta\Rightarrow |f(B'_{r}(a))-b|<\epsilon. Also, when x_{2}>0, |x_{2}-a|<\delta\Rightarrow |f(B'_{r}(a))-b|<\epsilon.

    The only way that this can happen is when b=0?????
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  4. #4
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    Quote Originally Posted by zebra2147 View Post
    Attempt #1 at proof:

    Suppose D=D_{1}\bigcup D_{2} where D_{1} contains only negative real numbers and D_{2} contains only positive real values. Let a be a limit point of D_{1} and D_{2}. Then let f: D\rightarrow R be the function f(B'_{r}(a)). Thus, the lim_{x\longrightarrow a} exists only if lim_{x\longrightarrow a}f|_D_{j} both exist and are equal.

    Then, if both limits exist we have that, for some \epsilon>0,when x_{1}<0, |x_{1}-a|<\delta\Rightarrow |f(B'_{r}(a))-b|<\epsilon. Also, when x_{2}>0, |x_{2}-a|<\delta\Rightarrow |f(B'_{r}(a))-b|<\epsilon.

    The only way that this can happen is when b=0?????

    I think you confused stuff here big time: the positive and negative numbers are in the image of f , NOT in its domain!

    We know that \forall\epsilon>0\,\exists\delta_\epsilon>0\,\,s.t  .\,\,x\in B_{\delta_\epsilon}(x_0)\Longrightarrow f(x)\in B_\epsilon(b).

    Suppose b>0 , and let us choose \epsilon=b/2 . By the above, we get that f(x)\in B_\epsilon(b) for any

    point x\in B_{\delta_\epsilon}(x_0) , for some \delta_\epsilon>0 ,or in other words: f\left(B_{\delta_\epsilon}(x_0)\right)\subset B_\epsilon(b) ...but this can't be since

    in f\left(B_{\delta_\epsilon}(x_0)\right) there's some negative number, whereas in B_\epsilon(b) all the numbers are positive...

    Read the above slowly and with a pencil and a piece of paper by your side...draw diagrams if that helps you.

    Tonio
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  5. #5
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    Oh ok. That makes more sense. Then I'm assuming that you showed how b>0 does not work, so if we show that b<0 does not work by a similar proof then b=0 must be true?
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  6. #6
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    Quote Originally Posted by zebra2147 View Post
    Oh ok. That makes more sense. Then I'm assuming that you showed how b>0 does not work, so if we show that b<0 does not work by a similar proof then b=0 must be true?

    Of course. I think you don't even need to show for b<0 since it is plain that this case is completely similar to the one already proven.

    Tonio
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