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Math Help - question about limits

  1. #1
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    question about limits

    if
    \lim_{x\to \infty}\frac{g(x)}{f(x)} = 1
    and
    \lim_{x\to \infty}\frac{h(x)}{f(x)} = 1
    then does this imply that
    \lim_{x\to \infty}\frac{g(x)}{h(x)} = 1 also?
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  2. #2
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    Quote Originally Posted by Aradesh
    if
    \lim_{x\to \infty}\frac{g(x)}{f(x)} = 1
    and
    \lim_{x\to \infty}\frac{h(x)}{f(x)} = 1
    then does this imply that
    \lim_{x\to \infty}\frac{g(x)}{h(x)} = 1 also?
    If \frac{h(x)}{f(x)}>0 then,
    \left(\lim_{x\to\infty}\frac{h(x)}{f(x)}\right)^{-1}=\lim_{x\to\infty}\frac{f(x)}{h(x)} because the function \frac{1}{x} is countinous for x>0. Since,
    \lim_{x\to\infty}\frac{h(x)}{f(x)}=1
    Then,
    \lim_{x\to\infty}\frac{f(x)}{h(x)}=1
    Then,
    \lim_{x\to\infty}\frac{f(x)}{h(x)}\lim_{x\to\infty  }\frac{g(x)}{f(x)}=1
    Since the limits exists,
    \lim_{x\to\infty}\frac{\not{f(x)}}{h(x)}\frac{g(x)  }{\not{f(x)}}=1

    Note: I only proved this for a specific case when that fraction is always positive. I do not know if it is true in a general case, but it might.
    Last edited by ThePerfectHacker; March 3rd 2006 at 01:01 PM.
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  3. #3
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    thanks.
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  4. #4
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    Quote Originally Posted by Aradesh
    thanks.
    Welcomed.
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