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Math Help - Limit of a floor function

  1. #1
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    Limit of a floor function

     \lim_ {x \rightarrow \00} (x[1/x])=?
    Hello,
    Can you please help me finding the limit above?
    the squared brackets mean floor function.

    Thanks,
    Michael
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  2. #2
    Super Member girdav's Avatar
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    We have \displaystyle \lim_{x\to 0 }x\lfloor \frac 1x\rfloor =\lim_{t\to\infty}\frac 1t\lfloor t\rfloor. Now you can use a well-known inequality.
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  3. #3
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    To make notation easier say f(x) = x\left\lfloor {\frac{1}{x}} \right\rfloor .
    Look at the limit  \displaystyle\lim _{x \to 0^ +  } f(x).
    For each n\in\mathbb{Z}^+ if x\in\left(\frac{1}{n+1},\frac{1}{n}\right) then n<\frac{1}{x}<n+1.
    So it follows that \left\lfloor {\frac{1}{x}} \right\rfloor={n} which implies that \frac{n}{{n + 1}} < f(x) < 1

    So what is the limit on the right?

    You can do the limit on the left.
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  4. #4
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    Thank a lot, but I cannot understand what are you trying to say, can you please simplify your answer?
    I remind you the the squared brackets mean floor(1/X)
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  5. #5
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    Quote Originally Posted by msokol89 View Post
    Thank a lot, but I cannot understand what are you trying to say, can you please simplify your answer?
    I remind you the the squared brackets mean floor(1/X)
    Do you know anything about the floor function?
    How does it work?

    BTW, \lfloor x\rfloor is the standard notation.
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  6. #6
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    Yes, it gives the highest integer that is lower than a specific number,
    My problem is that I am not that familiar with the official notations.
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  7. #7
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    Quote Originally Posted by msokol89 View Post
    Yes, it gives the highest integer that is lower than a specific number, My problem is that I am not that familiar with the official notations.
    Here is a web reference.
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