please help me! L=lim_(x→1) ((p/(1−x^p ))−(q/(1−x^q ))) , (p,q∈R)


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Question Number 46268 by Saorey last updated on 23/Oct/18

$please help me!$ $L = \lim_{x \to 1} (\frac{p}{1 - x^{p}} - \frac{q}{1 - x^{q}}), (p, q \in R)$
Commented by maxmathsup by imad last updated on 23/Oct/18

$l e t A (x) = \frac{p}{1 - x^{p}} - \frac{q}{1 - x^{q}} c h a n g e m e n t x = 1 + t g i v e$ $A (x) = B (t) \Rightarrow {l i m}_{x \to 1} A (x) = {l i m}_{t \to 0} B (t) w i t h B (t) = \frac{p}{1 - {(1 + t)}^{p}} - \frac{q}{1 - {(1 + t)}^{q}}$ $b u t {(1 + t)}^{p} \sim 1 + p t + \frac{p (p - 1)}{2} t^{2} a n d {(1 + t)}^{q} \sim 1 + q t + \frac{q (q - 1)}{t} t^{2} (t \to o) \Rightarrow$ $B (t) \sim \frac{p}{- p t - \frac{p (p - 1)}{2} t^{2}} - \frac{q}{- q t - \frac{q (q - 1)}{2} t^{2}} \Rightarrow$ $B (t) \sim - \frac{1}{t + \frac{p - 1}{2} t^{2}} + \frac{1}{t + \frac{(q - 1)}{2} t^{2}} = \frac{- t - \frac{(q - 1)}{2} t^{2} + t + \frac{p - 1}{2} t^{2}}{t^{2} (1 + \frac{(p - 1) t}{2}) (1 + \frac{q - 1}{2} t)}$ $= \frac{\frac{p - q}{2}}{(1 + \frac{p - 1}{2} t) (1 + \frac{q - 1}{2} t)} \to \frac{p - q}{2} (t \to 0) \Rightarrow L = {l i m}_{x \to 1} A (x) = \frac{p - 1}{2} .$
Commented by maxmathsup by imad last updated on 23/Oct/18

$L = {l i m}_{x \to 1} A (x) = \frac{p - q}{2} .$
	Answered by MJS last updated on 23/Oct/18

	$\lim_{x \to 1} \frac{p}{1 - x^{p}} - \frac{q}{1 - x^{q}} = \lim_{x \to 1} \frac{p (1 - x^{q}) - q (1 - x^{p})}{(1 - x^{p}) (1 - x^{q})} =$ $[l^{'} Hopital, 2 times because 1^{st} derivate$ $is still undefined]$ $= \lim_{x \to 1} \frac{\frac{d^{2}}{{d x}^{2}} [p (1 - x^{q}) - q (1 - x^{p})]}{\frac{d^{2}}{{d x}^{2}} [(1 - x^{p}) (1 - x^{q})]} =$ $= \lim_{x \to 1} \frac{p q (p - 1) x^{p - 2} + p q (1 - q) x^{q - 2}}{(p + q) (p + q - 1) x^{p + q - 2} - p (p - 1) x^{p - 2} - q (q - 1) x^{q - 2}} =$ $= \frac{p - q}{2}$
	Commented by Saorey last updated on 23/Oct/18

	$can you calculate it without lopital ?$
	Answered by ajfour last updated on 24/Oct/18
	$L=lim_(x→1) {(p/(1−[1−(1−x)]^p ))−(q/(1−[1−(1−x)]^q ))} = lim_(h→0) {(p/(1−(1−ph+((p(p−1)h^2 )/2)−..)))−(q/(1−(1−qh+((q(q−1)h^2 )/2)−..)))} ⇒ L = lim_(h→0) {(1/h)[(1/(1−(((p−1)h)/2)))−(1/(1−(((q−1)h)/2)))]} = lim_(h→0) (1/h)[(((1−(((q−1)h)/2))−(1−(((p−1)h)/2)))/((1−(((p−1)h)/2))(1−(((q−1)h)/2))))] =(((((p−q)/2)))/1) =((p−q)/2) .$
	$L = \lim_{x \to 1} {\frac{p}{1 - {[1 - (1 - x)]}^{p}} - \frac{q}{1 - {[1 - (1 - x)]}^{q}}}$ $= \lim_{h \to 0} {\frac{p}{1 - (1 - p h + \frac{p (p - 1) h^{2}}{2} - . .)} - \frac{q}{1 - (1 - q h + \frac{q (q - 1) h^{2}}{2} - . .)}}$ $\Rightarrow L = \lim_{h \to 0} {\frac{1}{h} [\frac{1}{1 - \frac{(p - 1) h}{2}} - \frac{1}{1 - \frac{(q - 1) h}{2}}]}$ $= \lim_{h \to 0} \frac{1}{h} [\frac{(1 - \frac{(q - 1) h}{2}) - (1 - \frac{(p - 1) h}{2})}{(1 - \frac{(p - 1) h}{2}) (1 - \frac{(q - 1) h}{2})}]$ $= \frac{(\frac{p - q}{2})}{1} = \frac{p - q}{2} .$
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