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Math Help - limits of a fraction

  1. #1
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    limits of a fraction

    how do you find the limits of [(1+x) ^(1/2) -1 ] / [ (1-x)^(1/2) -1 ]?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    how do you find the limits of [(1+x) ^(1/2) -1 ] / [ (1-x)^(1/2) -1 ]?
    x tends to?
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  3. #3
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    oh sorry! i missed that out..it is as x tends to 0.
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  4. #4
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    0/0 so use L'Hopital's Rule
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  5. #5
    Super Member Random Variable's Avatar
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     \displaystyle  = \lim_{x \to 0} \frac{\Big(1+\frac{x}{2}-\frac{x^{2}}{8} + O(x^3)\Big) - 1}{\Big(1- \frac{x}{2} - \frac{x^{2}}{8} + O(x^{3})\Big)- 1}

     \displaystyle = \lim_{x \to 0} \frac{\frac{1}{2} - \frac{x}{8} + O(x^{2})}{-\frac{1}{2} - \frac{x}{8} + O(x^{2})} = -1
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  6. #6
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    Quote Originally Posted by alexandrabel90 View Post
    how do you find the limits of [(1+x) ^(1/2) -1 ] / [ (1-x)^(1/2) -1 ]?
    If you multiply numerator and denominator by (1+x)^(1/2)+ 1 and multiply numerator and denominator by (1- x)^(1/2)+ 1 you get
    \left(\frac{(1-x)^{1/2}+1}{(1+x)^{1/2}+ 1}\right)\left(\frac{1+ x- 1}{1-x- 1}\right)
    = -\frac{(1-x)^{1/2}+ 1}{(1+x)^{1/2}+ 1}
    and now you can take x= 0.
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