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December 22nd, 2009, 04:02 AM  #1 
Newbie Joined: Dec 2009 Posts: 9 Thanks: 0  How to factorise the difference of noninteger powers?
I have a problem to find the limit as x goes to 2 of (x^1/32^1/3)/(x2) [that is the cube root of x minus the cube root of two, all divided by x minus 2]. As far as I know, to solve this, the numerator needs to be factorised into a factor of (x2) and another factor (a series?), just as you would do for integer powers. But here the powers are 1/3, so how do you find the second factor? I guess there are other ways to solve this also, but I am specifically interested in the method described above. Thanks. 
December 22nd, 2009, 05:01 AM  #2 
Senior Member Joined: Dec 2009 From: Las Vegas Posts: 209 Thanks: 0  Re: How to factorise the difference of noninteger powers?
Hi anonymoose; If you expand x^(1/3) with a taylor series around x = 2 you will get: equals: simplify: Now if you take the limit as x approaches two term by term you will get the answer. 
December 22nd, 2009, 06:57 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,835 Thanks: 2162  
December 22nd, 2009, 07:01 AM  #4 
Newbie Joined: Dec 2009 Posts: 9 Thanks: 0  Re: How to factorise the difference of noninteger powers?
Thanks for the clear explanations.


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