Question

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

Answer 1

\[\large \lim_{x\rightarrow 1}\frac{p}{1-x^{p}} - \frac{q}{1 - x^{q}}\]

Like that?

Answer 2

Yes

Answer 3

@pooja195

Answer 4

Hint: take common fact, add, then de l'Hôpital's rule, twice

Answer 5

@mathmate

Answer 6

It would not helping

Answer 7

@mathmate

Answer 8

using the hint of @mathmate I got this expression:
\[\frac{{ - pq\left( {q - 1} \right){x^{q - 2}} + qp\left( {p - 1} \right){x^{p - 2}}}}{{ - q\left( {q - 1} \right){x^{q - 2}} - p\left( {p - 1} \right){x^{p - 2}} + \left( {p + q} \right)\left( {p + q - 1} \right){x^{p + q - 2}}}}\]

Answer 9

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:\[L=\lim_{t \to 0} \left( \frac{p}{-pt-\frac{1}{2}(p-1)pt^2}-\frac{q}{-qt-\frac{1}{2}(q-1)qt^2}\right)\]\[L=\lim_{t \to 0} \left( \frac{1}{t}\left( \frac{1}{-1-\frac{1}{2}(p-1)t}-\frac{1}{-1-\frac{1}{2}(q-1)t}\right)\right) \]\[L=\lim_{t \to 0} \left( \frac{1}{t}\left( \frac{1}{1+\frac{1}{2}(q-1)t}-\frac{1}{1+\frac{1}{2}(p-1)t}\right)\right) \]\[L=\lim_{t \to 0} \left( \frac{1}{t} \frac{\frac{1}{2}(p-1)t-\frac{1}{2}(q-1)t}{\left(1+\frac{1}{2}(q-1)t\right)\left(1+\frac{1}{2}(p-1)t\right)}\right) \]\[L=\lim_{t \to 0} \left( \frac{\frac{1}{2}(p-1)-\frac{1}{2}(q-1)}{\left(1+\frac{1}{2}(q-1)t\right)\left(1+\frac{1}{2}(p-1)t\right)}\right) \]\[L=\frac{p-q}{2}\]

Answer 10

@mukushla brilliant!

Answer 11

Thanks @mathmate

Answer 12

@mukushla Could you please tell me the name of the expansion form? Much appreciate.

Answer 13

\[(1+t)^a=1+at+\frac{1}{2}(a-1)at^2+O(t^3)\]

Answer 14

I think it doesn't have a specific name, let me search

Answer 15

I saw \(O\) function before, but not in this expansion.

Answer 16

that's just taylor series :D

Answer 17

https://en.wikipedia.org/wiki/Big_O_notation

Answer 18

Thank you.

Answer 19

np