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Math Help - limit proof

  1. #1
    Super Member Random Variable's Avatar
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    limit proof

    In the following proof, why does the fact that  \lim_{x \to 0} f(x) =0 imply that  \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)| ?

    http://www.math.purdue.edu/pow/sprin.../solution9.pdf
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  2. #2
    Member mabruka's Avatar
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    One way of seeing it:

    Absolute value is a continuous function so \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|=\Big|\lim_{x \to 0} f(x) - \lim_{x \to 0} f\Big(\frac{x}{2^{n}}\Big) \Big| =|0-0|=0

    Where it was used the fact that since when x\to 0 then \frac{x}{2^n}\to 0 as well, we have
    \lim_{x \to 0} f\Big(\frac{x}{2^{n}}\big)=\lim_{\frac{x}{2^n} \to 0} f\Big(\frac{x}{2^{n}}\big)=\lim_{y \to 0} f(y)=0


    Finally

    \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|=0=\left|\lim_{x \to 0} f(x)\right |=\lim_{x \to 0} |f(x)|
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  3. #3
    Member mabruka's Avatar
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    Quote Originally Posted by mabruka View Post
    One way of seeing it:

    Absolute value is a continuous function so \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|=\Big|\lim_{x \to 0} f(x) - \lim_{x \to 0} f\Big(\frac{x}{2^{n}}\Big) \Big| =|0-0|=0

    Where it was used the fact that since when x\to 0 then \frac{x}{2^n}\to 0 as well, we have
    \lim_{x \to 0} f\Big(\frac{x}{2^{n}}\big)=\lim_{\frac{x}{2^n} \to 0} f\Big(\frac{x}{2^{n}}\big)=\lim_{y \to 0} f(y)=0


    Finally

    \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|=0=\left|\lim_{x \to 0} f(x)\right |=\lim_{x \to 0} |f(x)|

    Hmmm i totally missread the question, i am sorry let me check it again
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  4. #4
    Member mabruka's Avatar
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    Certainly i dont understand how that step is done,

    in fact it would seem to me that \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)| doesnt make any sense since the left side does not depend on x.

    Returning to the proof, earlier we had that

    \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|<2\epsilon x for all n if 0< x <\alpha

    Taking n\to\infty and using the continuity of the abs value we get that since if n\to \infty then y=\frac{x}{2^n}\to 0,

    in our case :
    if n\to \infty    then f(\frac{x}{2^n})=f(y)\to 0.

    And we end up having |f(x)|\leq 2\epsilon x  from where the proof follows quite easily.
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by mabruka View Post
    Certainly i dont understand how that step is done,

    in fact it would seem to me that \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)| doesnt make any sense since the left side does not depend on x.

    Returning to the proof, earlier we had that

    \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|<2\epsilon x for all n if 0< x <\alpha

    Taking n\to\infty and using the continuity of the abs value we get that since if n\to \infty then y=\frac{x}{2^n}\to 0,

    in our case :
    if n\to \infty    then f(\frac{x}{2^n})=f(y)\to 0.

    And we end up having |f(x)|\leq 2\epsilon x  from where the proof follows quite easily.
    EDIT : So you think it's just a typo and what they're really doing is taking the limit as n approaches infinity?
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  6. #6
    Member mabruka's Avatar
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    Either a typo or some argument (which is not clear, at least for me) equivalent to the n-limit one. So far i dont see any flaws on this one.
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  7. #7
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    Quote Originally Posted by Random Variable View Post
    In the following proof, why does the fact that  \lim_{x \to 0} f(x) =0 imply that  \lim_{x \to 0} \Big|f(x) - f\Big(\frac{x}{2^{n}}\Big) \Big|= |f(x)| ?

    http://www.math.purdue.edu/pow/sprin.../solution9.pdf
    It certainly looks to me like this was supposed to be
    \lim_{n\to \infty}\left|f(x)- f(x/2^n)\right|= |f(x)|
    Because the right hand side depends on x, not n. But, since that is \left|f(x)- \lim_{n\to\infty}f(x/2^n)\right| it looks obvious.
    Last edited by Plato; May 23rd 2010 at 05:16 AM. Reason: LaTeX
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