how do you find the limits of [(1+x) ^(1/2) -1 ] / [ (1-x)^(1/2) -1 ]?

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- Jan 9th 2011, 04:48 PMalexandrabel90limits of a fraction
how do you find the limits of [(1+x) ^(1/2) -1 ] / [ (1-x)^(1/2) -1 ]?

- Jan 9th 2011, 04:50 PMAlso sprach Zarathustra
- Jan 9th 2011, 05:15 PMalexandrabel90
oh sorry! i missed that out..it is as x tends to 0.

- Jan 9th 2011, 05:16 PMdwsmith
0/0 so use L'Hopital's Rule

- Jan 9th 2011, 06:26 PMRandom Variable
$\displaystyle \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} $

$\displaystyle \displaystyle = \lim_{x \to 0} \frac{\frac{1}{2} - \frac{x}{8} + O(x^{2})}{-\frac{1}{2} - \frac{x}{8} + O(x^{2})} = -1$ - Jan 10th 2011, 01:26 PMHallsofIvy
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

$\displaystyle \left(\frac{(1-x)^{1/2}+1}{(1+x)^{1/2}+ 1}\right)\left(\frac{1+ x- 1}{1-x- 1}\right)$

$\displaystyle = -\frac{(1-x)^{1/2}+ 1}{(1+x)^{1/2}+ 1}$

and now you can take x= 0.