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$$\lim_{x\to 1}\frac9{1-x^9}-\frac7{1-x^7}$$

I did this with l'Hôpital's rule, but how can we do this problem other than that? Any hints will be good.

Did
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user354545
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    $1-x^n = (1-x)\displaystyle\sum_{i=0}^{n-1}x^i$ is another way to go – Maxime Ramzi Jul 07 '17 at 09:01
  • Let $x=1+t$ then $t\to0$ when $x\to1$ hence $$1-x^9=1-(1+t)^9=-9t-36t^2+o(t^2)$$ and $$1-x^7=1-(1+t)^7=-7t-21t^2+o(t^2)$$ which leads to $$f(x)=\frac{9}{1-x^9}-\frac{7}{1-x^7}=-\frac1{t+4t^2+o(t^2)}-\frac1{t+3t^2+o(t^2)}$$ that is, $$f(x)=\frac1t\left(\frac1{1+4t+o(t)}-\frac1{1+3t+o(t)}\right)$$ and finally, $$f(x)=\frac1t\left(1-4t-(1-3t)+o(t)\right)=-1+o(1)$$ – Did Jul 07 '17 at 09:07
  • Or: $$2-9x^7+7x^9=(1-x)^2(2+4x+6x^2+8x^3+10x^4+12x^5+14x^6+7x^7)\sim63(1-x)^2$$ – Did Jul 07 '17 at 09:16
  • For the record, I forgot a minus sign in the "that is" step of my first comment hence the limit is $+1$, not $-1$. – Did Jul 07 '17 at 09:19
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    @Did: If I were you, I'd write up your comments as an answer, so you can edit it however you want. – Theo Bendit Jul 07 '17 at 09:20
  • And finally, the same approach yields, for every $(a,b)$, $$\lim_{x\to1}\frac{a}{1-x^a}-\frac{b}{1-x^b}=\frac{a-b}2$$ – Did Jul 07 '17 at 09:22
  • @Did thanks got it :) – user354545 Jul 07 '17 at 09:23

2 Answers2

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The difference of fractions equals $\frac{9(1-x^7) - 7(1-x^9)}{(1-x^7)(1-x^9)}$.

Try to divide out the common $(1-x)^2$ terms in both the bottom and the top.

In the bottom we get $(1+x+x^2+\ldots+x^8)(1+x + x^2 + \ldots + x^6)$

In the top $7x^7 + 14x^6 + 12x^5 + 10x^4 + 8x^3 + 6x^2 + 4x + 2$ (multiply by $(1-x)^2$ to check both). Now we can just substitute $x=1$ in both. I get 63 twice OTTOMH.

Henno Brandsma
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I think the problem is

$$\lim_{x\to 1}\frac{9}{1-x^9}-\frac{7}{1-x^7}$$

Set $x=1-y$

$$\dfrac9{1-x^9}-\dfrac7{1-x^7}=\cdots=\dfrac1{y-4y^2+O(y^3)}-\dfrac1{y-3y^2+O(y^3)}=?$$

lab bhattacharjee
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