limits of a fraction

**alexandrabel90** · January 9th 2011, 04:48 PM

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

**Also sprach Zarathustra** · January 9th 2011, 04:50 PM

Originally Posted by alexandrabel90

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

x tends to?

**alexandrabel90** · January 9th 2011, 05:15 PM

oh sorry! i missed that out..it is as x tends to 0.

**dwsmith** · January 9th 2011, 05:16 PM

0/0 so use L'Hopital's Rule

**Random Variable** · January 9th 2011, 06:26 PM

$\displaystyle = \lim_{x \to 0} \frac{\Big(1+\frac{x}{2}-\frac{x^{2}}{8} + O(x^3)\Big) - 1}{\Big(1- \frac{x}{2} - \frac{x^{2}}{8} + O(x^{3})\Big)- 1}$

$\displaystyle = \lim_{x \to 0} \frac{\frac{1}{2} - \frac{x}{8} + O(x^{2})}{-\frac{1}{2} - \frac{x}{8} + O(x^{2})} = -1$

**HallsofIvy** · January 10th 2011, 01:26 PM

Originally Posted by alexandrabel90

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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