Let $m,n\in\mathbb N$. How can I solve this limit without using l'Hopital's rule, please?
$$\lim_{x\to1}\left(\frac{m}{1-x^m}-\frac{n}{1-x^n}\right)$$
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2Let $x=y+1$ and use the binomial theorem. – Claude Leibovici Nov 05 '17 at 10:12
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2Some non L'Hopital answers are here Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ as well as here Find the limit $\lim_{x \to 1} \left(\frac{p}{1-x^p} - \frac{q}{1-x^q}\right) $ $p ,q >0$. – Sil Nov 05 '17 at 10:22
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\begin{align}\lim_{x\to1}\left(\frac m{x^m-1}-\frac n{x^n-1}\right)&=\lim_{x\to1}\frac{m(x^n-1)-n(x^m-1)}{(x^m-1)(x^n-1)}\\&=\lim_{x\to1}\frac{m(x-1)(x^{n-1}+\cdots+1)-n(x-1)(x^{m-1}+\cdots+1)}{(x^m-1)(x^n-1)}\\&=\lim_{x\to1}\frac{m(x^{n-1}+\cdots+1)-n(x^{m-1}+\cdots+1)}{(x-1)(x^{m-1}+\cdots+1)(x^{n-1}+\cdots+1)}\\&=\frac1{mn}\lim_{x\to1}\frac{m(x^{n-1}+\cdots+1)-n(x^{m-1}+\cdots+1)}{x-1}.\end{align}By definition of derivative, the previous limit is the derivative at $1$ of the function$$x\mapsto m(x^{n-1}+x^{n-2}+\cdots+1)-n(x^{m-1}+x^{m-2}+\cdots+1),$$which is\begin{align}m\frac{(n-1)n}2-n\frac{(m-1)m}2&=\frac{mn^2-mn-nm^2+nm}2\\&=\frac{mn(n-m)}2.\end{align}Therefore, your limit is $\dfrac{n-m}2$.
José Carlos Santos
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