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Pawanyadav
 one year ago
Evaluate. lim{p/1x^p  q/1x^q}
X>1
Pawanyadav
 one year ago
Evaluate. lim{p/1x^p  q/1x^q} X>1

This Question is Closed

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \lim_{x\rightarrow 1}\frac{p}{1x^{p}}  \frac{q}{1  x^{q}}\] Like that?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4Hint: take common fact, add, then de l'Hôpital's rule, twice

Pawanyadav
 one year ago
Best ResponseYou've already chosen the best response.1It would not helping

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1using the hint of @mathmate I got this expression: \[\frac{{  pq\left( {q  1} \right){x^{q  2}} + qp\left( {p  1} \right){x^{p  2}}}}{{  q\left( {q  1} \right){x^{q  2}}  p\left( {p  1} \right){x^{p  2}} + \left( {p + q} \right)\left( {p + q  1} \right){x^{p + q  2}}}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sure there are some cool methods for solving this limit problem, I had an immediate thinking and came up with a series expansion approach, which I write for you here: let \(x=1+t\) with \(t \to 0\) It follows that \[L=\lim_{t \to 0} \left( \frac{p}{1(1+t)^p}\frac{q}{1(1+t)^q}\right)\]and with series expansion\[(1+t)^a=1+at+\frac{1}{2}(a1)at^2+O(t^3)\]Neglecting \(O(t^3)\) and and writing series expansion for \((1+t)^p\) and \((1+t)^q\) gives us:\[L=\lim_{t \to 0} \left( \frac{p}{pt\frac{1}{2}(p1)pt^2}\frac{q}{qt\frac{1}{2}(q1)qt^2}\right)\]\[L=\lim_{t \to 0} \left( \frac{1}{t}\left( \frac{1}{1\frac{1}{2}(p1)t}\frac{1}{1\frac{1}{2}(q1)t}\right)\right) \]\[L=\lim_{t \to 0} \left( \frac{1}{t}\left( \frac{1}{1+\frac{1}{2}(q1)t}\frac{1}{1+\frac{1}{2}(p1)t}\right)\right) \]\[L=\lim_{t \to 0} \left( \frac{1}{t} \frac{\frac{1}{2}(p1)t\frac{1}{2}(q1)t}{\left(1+\frac{1}{2}(q1)t\right)\left(1+\frac{1}{2}(p1)t\right)}\right) \]\[L=\lim_{t \to 0} \left( \frac{\frac{1}{2}(p1)\frac{1}{2}(q1)}{\left(1+\frac{1}{2}(q1)t\right)\left(1+\frac{1}{2}(p1)t\right)}\right) \]\[L=\frac{pq}{2}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@mukushla Could you please tell me the name of the expansion form? Much appreciate.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1\[(1+t)^a=1+at+\frac{1}{2}(a1)at^2+O(t^3)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it doesn't have a specific name, let me search

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I saw \(O\) function before, but not in this expansion.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's just taylor series :D
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