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Question Number 36024 by bshahid010@gmail.com last updated on 27/May/18

Commented by abdo mathsup 649 cc last updated on 27/May/18
$let use the changement x =1+t so (m/(x^m −1)) −(p/(x^p −1)) =(m/((1+t)^m −1)) − (p/((1+t)^p −1)) but (1+t)^m ∼ 1+mt +((m(m−1))/2) t^2 (1+t)^p ∼ 1+pt +((p(p−1))/2) t^2 ⇒ (m/(x^m −1)) −(p/(x^p −1)) ∼ (m/(mt+((m(m−1))/2)t^2 )) −(p/(pt +((p(p−1))/2)t^2 )) = (1/(t +((m−1)/2)t^2 )) − (1/(t +((p−1)/2)t^2 )) =2{ (1/(2t +(m−1)t^2 )) −(1/(2t +(p−1)t^2 ))} =2{ ((2t +(p−1)t^2 −2t −(m−1)t^2 )/(t^2 (2 +(m−1)t )(2+(p−1)t)))} = 2 ((p−m)/((2 +(m−1)t)( 2+(p−1)t))) →_(t→0) ((p−m)/2)$
$l e t u s e t h e c h a n g e m e n t x = 1 + t s o$ $\frac{m}{x^{m} - 1} - \frac{p}{x^{p} - 1} = \frac{m}{{(1 + t)}^{m} - 1} - \frac{p}{{(1 + t)}^{p} - 1} b u t$ ${(1 + t)}^{m} \sim 1 + m t + \frac{m (m - 1)}{2} t^{2}$ ${(1 + t)}^{p} \sim 1 + p t + \frac{p (p - 1)}{2} t^{2} \Rightarrow$ $\frac{m}{x^{m} - 1} - \frac{p}{x^{p} - 1} \sim \frac{m}{m t + \frac{m (m - 1)}{2} t^{2}} - \frac{p}{p t + \frac{p (p - 1)}{2} t^{2}}$ $= \frac{1}{t + \frac{m - 1}{2} t^{2}} - \frac{1}{t + \frac{p - 1}{2} t^{2}}$ $= 2 {\frac{1}{2 t + (m - 1) t^{2}} - \frac{1}{2 t + (p - 1) t^{2}}}$ $= 2 {\frac{2 t + (p - 1) t^{2} - 2 t - (m - 1) t^{2}}{t^{2} (2 + (m - 1) t) (2 + (p - 1) t)}}$ $= 2 \frac{p - m}{(2 + (m - 1) t) (2 + (p - 1) t)} \to_{t \to 0} \frac{p - m}{2}$
Commented by tanmay.chaudhury50@gmail.com last updated on 27/May/18

$e x c e l l e n t . . .$
	Answered by ajfour last updated on 27/May/18

	$= \lim_{x \to 1} (\frac{m}{m h + \frac{m (m - 1)}{2} h^{2}} - \frac{p}{p h + \frac{p (p - 1)}{2} h^{2}})$ $= \lim_{x \to 1} (\frac{\frac{(p - 1)}{2} h^{2} - \frac{(m - 1)}{2} h^{2} + . . .}{h^{2} + * h^{3} + . . .})$ $= \frac{p - m}{2} .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 27/May/18

	$g o o d . . . h e r e y o u h a v e i g n o r e d h i g h e r p o w e r o f$ $h a n d h = x - 1$
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