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December 22nd, 2009, 04:02 AM   #1
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How to factorise the difference of non-integer powers?

I have a problem to find the limit as x goes to 2 of (x^1/3-2^1/3)/(x-2) [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 (x-2) 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.
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December 22nd, 2009, 05:01 AM   #2
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Re: How to factorise the difference of non-integer 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.

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December 22nd, 2009, 06:57 AM   #3
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December 22nd, 2009, 07:01 AM   #4
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Re: How to factorise the difference of non-integer powers?

Thanks for the clear explanations.
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