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Math Help - Limit Problem

  1. #1
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    Smile Limit Problem

     lim_{x \to 0^{+}} \frac{b}{x}*[\frac{x}{a}] =

    (a > 0 , b > 0)
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  2. #2
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    I must be reading your question wrong, because it looks like you have

     \mathop{\lim}\limits_{x \to 0+} \frac{b}{x} * \frac{x}{a}

    In which case those x's will cancel and the limit is just

     \frac{b}{a}
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  3. #3
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    No , it's with [ ]
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  4. #4
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    Which is NOT standard notation. I would guess that you mean the floor function which would be [x] without the upper serifs.
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  5. #5
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    Quote Originally Posted by gilyos View Post
    No , it's with [ ]
    Which is the greatest integer function or the floor function.
    We know that a>0 so it follows that \lim _{x \to 0^ +  } \frac{x}{a} = 0.
    That being the case at some point, \frac{x}{a}<1 so that means \left\lfloor {\frac{x}{a}} \right\rfloor  = 0.

    Now what can you say about this limit?
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  6. #6
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     \frac{x}{a}-1 <= [\frac{x}{a}] <= \frac{x}{a}

     (\frac{x}{a}-1)*\frac{b}{x} <= [\frac{x}{a}] <= \frac{x}{a}*\frac{b}{x} = \frac{b}{a}

    but , what I doing now ???
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  7. #7
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    Quote Originally Posted by gilyos View Post
     \frac{x}{a}-1 <= [\frac{x}{a}] <= \frac{x}{a}
     (\frac{x}{a}-1)*\frac{b}{x} <= [\frac{x}{a}] <= \frac{x}{a}*\frac{b}{x} = \frac{b}{a}
    but , what I doing now ???
    That has absolutely nothing to do with this question.
    Do you understand that \lim _{x \to 0^ +  } \frac{x}{a} = 0 means that \left( {\exists \delta  > 0} \right)\left[ {0 < x < \delta  \Rightarrow \quad 0<\frac{x}{a} < 1} \right]
    Which means in turn that \left( {\exists \delta  > 0} \right)\left[ {0 < x < \delta  \Rightarrow \quad \left\lfloor {\frac{x}{a}} \right\rfloor  = 0} \right]
    Which implies \left( {\exists \delta  > 0} \right)\left[ {0 < x < \delta  \Rightarrow \quad \frac{b}{x}\left\lfloor {\frac{x}{a}} \right\rfloor  = 0} \right].
    Note is does not matter what b is.
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