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Thread: Limits

  1. #1
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    Limits

    1.$\displaystyle
    Suppose \mid f(x) \mid \leq \mid \frac{B}{x} \mid for\ all\ x\ not\ equal\ to\ 0.$
    $\displaystyle Show\ that:
    \lim_{x \to 0} ~ x^2f(x)=0
    $

    2.$\displaystyle
    Suppose\ that\ \lim_{x \to 0} ~ f(x)=0\ and\ that\ \forall x \in R\ g(x) \leq B.$
    $\displaystyle Show \lim_{x \to 0} ~ f(x)g(x)=0
    $

    Sorry, I don't even know where to start..
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  2. #2
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    Quote Originally Posted by Pisces View Post
    1.$\displaystyle
    Suppose \mid f(x) \mid \leq \mid \frac{B}{x} \mid for\ all\ x\ not\ equal\ to\ 0.$
    $\displaystyle Show\ that:
    \lim_{x \to 0} ~ x^2f(x)=0
    $
    Note, $\displaystyle |x^2f(x)| \leq x^2\left| \frac{B}{x} \right| = B|x|$.
    But $\displaystyle \lim_{x\to 0}B|x| = 0$.
    Thus, by sequeeze theorem we have $\displaystyle \lim_{x\to 0}x^2f(x) = 0$.

    2.$\displaystyle
    Suppose\ that\ \lim_{x \to 0} ~ f(x)=0\ and\ that\ \forall x \in R\ g(x) \leq B.$
    $\displaystyle Show \lim_{x \to 0} ~ f(x)g(x)=0
    $
    This is similar. Try it.
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  3. #3
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    wow, that looks so simple..

    so does this look about right?

    $\displaystyle
    \forall x \in R\ g(x)f(x) \leq Bf(x).$
    $\displaystyle So\ lim_{x \to 0} ~ Bf(x)=0$
    $\displaystyle
    Thus,\ lim_{x \to 0} ~ f(x)g(x)=0$
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Pisces View Post
    wow, that looks so simple..

    so does this look about right?

    $\displaystyle
    \forall x \in R\ g(x)f(x) \leq Bf(x).$
    $\displaystyle So\ lim_{x \to 0} ~ Bf(x)=0$
    $\displaystyle
    Thus,\ lim_{x \to 0} ~ f(x)g(x)=0$
    not exactly. with that inequality, it would be best for you to use the squeeze theorem here, which you did not

    as TPH did, consider $\displaystyle |g(x)f(x)|$
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  5. #5
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    I'm confused.

    Could you give me a hint?

    edit: im not even sure i understand how he did the first one.
    Last edited by Pisces; Sep 1st 2008 at 05:35 PM.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Pisces View Post
    I'm confused.

    Could you give me a hint?

    edit: im not even sure i understand how he did the first one.
    $\displaystyle |g(x)f(x)| \le |Bf(x)| = B|f(x)|$

    now take the limit of that system


    you should look up the squeeze theorem
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  7. #7
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    I see the theorem, and I understand h(x)<=f(x)<=g(x) and if the limit of h(x)=0 and limit of g(x)=0 then the limit of f(x)=0..

    but all this other stuff isn't really clicking.

    edit: plz help I am really confused.

    is that not what I did?

    $\displaystyle
    lim_{x \to 0} ~ Bf(x)=0$
    $\displaystyle so\ lim_{x \to 0} ~ f(x)g(x)=0
    $
    Last edited by Pisces; Sep 1st 2008 at 07:10 PM.
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  8. #8
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Pisces View Post
    2.$\displaystyle
    Suppose\ that\ \lim_{x \to 0} ~ f(x)=0\ and\ that\ \forall x \in R\ g(x) \leq B.$
    $\displaystyle Show \lim_{x \to 0} ~ f(x)g(x)=0
    $
    I think this is wrong unless we have for all real number $\displaystyle x$, $\displaystyle {\color{red}|}g(x){\color{red}|}\leq B$.
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