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$$\lim_{x\to 1}\left(\frac{p}{1-x^p}-\frac{q}{1-x^q}\right) , p,q\in N$$

I tried fitting the formula $\lim_{x\to a}\frac{x^n - a^n}{x-a} = na^{n-1}$ but could not proceed further. Also I don't think that we should directly apply L'Hôpital's rule as it will be too lengthy.

mathreadler
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Sourav Suman
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1 Answers1

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Using Taylor expansions around $0$, namely:

  • $(1+u)^\alpha = 1+u+\frac{\alpha(\alpha-1)}{2}u^2 + o(u^2)$

  • $\frac{1}{1+u} = 1-u+o(u)$ (a specific case of the first, for $\alpha=-1$)

You have, setting $h\stackrel{\rm def}{=} 1-x \xrightarrow[x\to1]{}0$ (to center things around $0$, with which I am much more familiar than around $1$): $$\begin{align} \frac{p}{1-x^p}-\frac{q}{1-x^q} &= \frac{p}{1-(1-h)^p}-\frac{q}{1-(1-h)^q} \\ &= \frac{p}{1-(1-ph+\frac{p(p-1)}{2}h^2+o(h^2))}-\frac{q}{1-(1-qh+\frac{q(q-1)}{2}h^2+o(h^2))}\\ &= \frac{p}{ph-\frac{p(p-1)}{2}h^2+o(h^2)}-\frac{q}{qh-\frac{q(q-1)}{2}h^2+o(h^2)}\\ &= \frac{1}{h-\frac{p-1}{2}h^2+o(h^2)}-\frac{1}{h-\frac{q-1}{2}h^2+o(h^2)}\\ &= \frac{1}{h}\left(\frac{1}{1-\frac{p-1}{2}h+o(h)}-\frac{1}{1-\frac{q-1}{2}h+o(h)}\right)\\ &= \frac{1}{h}\left(1+\frac{p-1}{2}h+o(h)-(1+\frac{q-1}{2}h+o(h))\right)\\ &= \frac{1}{h}\left(\frac{p-1-(q-1)}{2}h+o(h)\right)\\ &= \frac{p-q}{2}+o(1) \xrightarrow[h\to0]{}\boxed{\frac{p-q}{2}} \end{align}$$

Clement C.
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    To the OP: N.B. that you have to keep the $h^2$ terms because both the constant and the first order terms cancel. BTW, that means that you'd have to go to second derivatives for L'Hôpital: the ratio of the first derivatives would still be indeterminate. – NickD Apr 26 '17 at 02:45
  • @Nick Indeed. For the OP: try to do it stopping at the $o(h)$ order in the second line. You'll see that if you do so, in the 5th and 6th line you'll end up with $\frac{1}{h}(1+o(1)-(1+o(1)) = \frac{1}{h}\cdot o(1)$, from which you cannot conclude anything -- so you need to go to order $o(h^2)$ in the first Taylor expansion. – Clement C. Apr 26 '17 at 12:31