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

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1. johnweldon1993

$\large \lim_{x\rightarrow 1}\frac{p}{1-x^{p}} - \frac{q}{1 - x^{q}}$ Like that?

2. Pawanyadav

Yes

3. Pawanyadav

@pooja195

4. mathmate

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

5. Pawanyadav

@mathmate

6. Pawanyadav

It would not helping

7. Pawanyadav

@mathmate

8. Michele_Laino

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

9. anonymous

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}$

10. mathmate

@mukushla brilliant!

11. anonymous

Thanks @mathmate

12. Loser66

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

13. Loser66

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

14. anonymous

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

15. Loser66

I saw $$O$$ function before, but not in this expansion.

16. anonymous

that's just taylor series :D

17. anonymous
18. Loser66

Thank you.

19. anonymous

np

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