Evaluate. lim{p/1-x^p - q/1-x^q} X->1

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Evaluate. lim{p/1-x^p - q/1-x^q} X->1

Mathematics
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\[\large \lim_{x\rightarrow 1}\frac{p}{1-x^{p}} - \frac{q}{1 - x^{q}}\] Like that?
Yes

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Hint: take common fact, add, then de l'Hôpital's rule, twice
It would not helping
using the hint of @mathmate I got this expression: \[\frac{{ - pq\left( {q - 1} \right){x^{q - 2}} + qp\left( {p - 1} \right){x^{p - 2}}}}{{ - q\left( {q - 1} \right){x^{q - 2}} - p\left( {p - 1} \right){x^{p - 2}} + \left( {p + q} \right)\left( {p + q - 1} \right){x^{p + q - 2}}}}\]
I'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\[(1+t)^a=1+at+\frac{1}{2}(a-1)at^2+O(t^3)\]Neglecting \(O(t^3)\) and and writing series expansion for \((1+t)^p\) and \((1+t)^q\) gives us:\[L=\lim_{t \to 0} \left( \frac{p}{-pt-\frac{1}{2}(p-1)pt^2}-\frac{q}{-qt-\frac{1}{2}(q-1)qt^2}\right)\]\[L=\lim_{t \to 0} \left( \frac{1}{t}\left( \frac{1}{-1-\frac{1}{2}(p-1)t}-\frac{1}{-1-\frac{1}{2}(q-1)t}\right)\right) \]\[L=\lim_{t \to 0} \left( \frac{1}{t}\left( \frac{1}{1+\frac{1}{2}(q-1)t}-\frac{1}{1+\frac{1}{2}(p-1)t}\right)\right) \]\[L=\lim_{t \to 0} \left( \frac{1}{t} \frac{\frac{1}{2}(p-1)t-\frac{1}{2}(q-1)t}{\left(1+\frac{1}{2}(q-1)t\right)\left(1+\frac{1}{2}(p-1)t\right)}\right) \]\[L=\lim_{t \to 0} \left( \frac{\frac{1}{2}(p-1)-\frac{1}{2}(q-1)}{\left(1+\frac{1}{2}(q-1)t\right)\left(1+\frac{1}{2}(p-1)t\right)}\right) \]\[L=\frac{p-q}{2}\]
@mukushla brilliant!
Thanks @mathmate
@mukushla Could you please tell me the name of the expansion form? Much appreciate.
\[(1+t)^a=1+at+\frac{1}{2}(a-1)at^2+O(t^3)\]
I think it doesn't have a specific name, let me search
I saw \(O\) function before, but not in this expansion.
that's just taylor series :D
https://en.wikipedia.org/wiki/Big_O_notation
Thank you.
np

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