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08-28-2008, 09:17 PM
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08-28-2008, 09:20 PM
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Quote:
Originally Posted by JOhkonut  \lim_x\to 2 \frac {2^\frac {x}{2} -2}{2^x-4}
Help! | Is this what you mean?  ??
--Chris
__________________ ![\color{blue}\text{ if }\Re[z]<0\text{ and }0<a\leq 1,](http://www.mathhelpforum.com/math-help/latex2/img/7997c8ff6f2bb4fa80412931e6ab2f1f-1.gif "\color{blue}\text{ if }\Re[z]<0\text{ and }0<a\leq 1,") ![\color{blue}\zeta(z,a)=\frac{2\Gamma(1-z)}{(2\pi)^{1-z}}\left[\sin\left(\frac{\pi z}{2}\right)\sum_{n=1}^{\infty}\frac{\cos(2\pi a n)}{n^{1-z}}+\cos\left(\frac{\pi z}{2}\right)\sum_{n=1}^{\infty}\frac{\sin(2\pi a n)}{n^{1-z}}\right]](http://www.mathhelpforum.com/math-help/latex2/img/2a5294e653c20ce63d99502635321876-1.gif "\color{blue}\zeta(z,a)=\frac{2\Gamma(1-z)}{(2\pi)^{1-z}}\left[\sin\left(\frac{\pi z}{2}\right)\sum_{n=1}^{\infty}\frac{\cos(2\pi a n)}{n^{1-z}}+\cos\left(\frac{\pi z}{2}\right)\sum_{n=1}^{\infty}\frac{\sin(2\pi a n)}{n^{1-z}}\right]") | | The following users thank Chris L T521 for this useful post: | |  |
08-28-2008, 09:21 PM
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Quote:
Originally Posted by Chris L T521 Is this what you mean? ??
--Chris | whoops, hehe I always seem to forget that. |
08-28-2008, 09:26 PM
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Quote:
Originally Posted by JOhkonut whoops, hehe I always seem to forget that. | hint: the denominator is the difference of two squares
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08-28-2008, 09:30 PM
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Quote:
Originally Posted by Jhevon hint: the denominator is the difference of two squares | Jhevon, ur great.  |
08-28-2008, 09:30 PM
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Quote:
Originally Posted by Jhevon hint: the denominator is the difference of two squares | ahahaha...why didn't I think of that? 
--Chris
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08-28-2008, 09:36 PM
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i believe with limits, if you come to something like
infinity/zero,
infinity*zero,
or zero/zero
you can take the derivative of the top and bottom expressions
and take the limit of those two expressions. i tried it out and they do simplify to a reasonable answer.
i'm not positive about all that info in there, the rule i'm refering to is called
l 'hospital it's french, try looking it up if you're still unsure
Last edited by potato salad123; 08-28-2008 at 09:49 PM.
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08-28-2008, 09:38 PM
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Quote:
Originally Posted by potato salad123 i believe with limits, if you come to something like
infinity*infinity,
infinity*zero,
or zero/zero
you can take the derivative of the top and bottom expressions
and take the limit of those two expressions. i tried it out and they do simplify to a reasonable answer. | I'm not supposed to know l'Hopital's rule yet |
08-28-2008, 09:56 PM
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