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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 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.
 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 non-integer powers? Hi anonymoose; If you expand x^(1/3) with a taylor series around x = 2 you will get: $\frac{\left (\sqrt[3]{2}\ + \ \frac{x-2}{3 \ 2^{2/3}} \ - \ \frac{(x-2)^2}{18 \ 2^{2/3}} \ + \ \frac{5 (x-2)^3}{324 \ 2^{2/3}} \ - \ \frac{5 (x-2)^4}{972 \ 2^{2/3}} \ + \ \frac{11 (x-2)^5}{5832 \ 2^{2/3}}- \ \dots + \right ) \ - \sqrt[3]{2}}{x-2}$ equals: $\frac{\left ( \ \frac{x-2}{3 \ 2^{2/3}} \ - \ \frac{(x-2)^2}{18 \ 2^{2/3}} \ + \ \frac{5 (x-2)^3}{324 \ 2^{2/3}} \ - \ \frac{5 (x-2)^4}{972 \ 2^{2/3}} \ + \ \frac{11 (x-2)^5}{5832 \ 2^{2/3}}- \ \dots + \right )}{x-2}$ simplify: $\left ( \ \frac{1}{3 \ 2^{2/3}} \ - \ \frac{(x-2)}{18 \ 2^{2/3}} \ + \ \frac{5 (x-2)^2}{324 \ 2^{2/3}} \ - \ \frac{5 (x-2)^3}{972 \ 2^{2/3}} \ + \ \frac{11 (x-2)^4}{5832 \ 2^{2/3}}- \ \dots + \right )$ Now if you take the limit as x approaches two term by term you will get the answer. $\lim_{x\to2}\frac{x^{1/3}-2^{1/3}}{x-2}=\left ( \frac{1}{3 \ 2^{2/3}} \right )$
 December 22nd, 2009, 06:57 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,835 Thanks: 2162 $\frac{x^{\frac13}\,-\,2^{\frac13}}{x\,-\,2}\,=\,\frac{x^{\frac13}\,-\,2^{\frac13}}{\left(x^{\frac13}\,-\,2^{\frac13}\right)\left(x^{\frac23}\,+\,2^{\frac 13}x^{\frac13}\,+\,2^{\frac23}\right)}\,=\,\frac{1 }{x^{\frac23}\,+\,2^{\frac13}x^{\frac13}\,+\,2^{\f rac23}}\,\to\,\frac{1}{3\left(2^{\frac23}\right)}\ text{ as }x\,\to\,2.$
 December 22nd, 2009, 07:01 AM #4 Newbie   Joined: Dec 2009 Posts: 9 Thanks: 0 Re: How to factorise the difference of non-integer powers? Thanks for the clear explanations.

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