limits.

[midnight jay]

I need an approach without using l hospital method

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[midnight jay]

@harsh plume

[harsh plume]

Alright, let's see.
First, let x = t + 1, t -> 0. Then we have:
p/(1 - (1 + t)^p) - q/(1 - (1 + t)^q), t -> 0
Then recall the expansion of (1 + x)^n.

Really, p and q don't even have to be natural numbers, they can be any real numbers.

[midnight jay]

Ok I see getting a similar term In terms of t

A term similar to the one given in question

Is that correct?

[harsh plume]

What do you mean?

[midnight jay]

Or can u use the standard formula Lim x-->1 (x^n - 1 )/(x-1) = nx^(n-1)

[harsh plume]

Oh, the definition of derivative? Well, it's pretty much the same as what I said, just with only one term.

[midnight jay]

I am saying that u can split the thing into two parts and one of them yields a term similar to the one given in the question but in terms of t. Since you said that p and q can be any real numbers, I can proceed by assuming it the term we started with

[harsh plume]

Well, alright. So, what do you want to do next?

[midnight jay]

With my plan after seeing ur idea, I am getting the correct answer and I discovered a new approach. But for now I would love to know how you would have proceeded

Lemme send solution after u explain

[harsh plume]

Alright.
p/(1 - (1 + t)^p) - q/(1 - (1 + t)^q), t -> 0
We have:
1 - (1 + t)^p = 1 - (1 + pt + p(p - 1)t^2/2 + o(t^2)) = -pt - p(p - 1)t^2 + o(t^2), t -> 0
p/(1 - (1 + t)^p) = p/(-pt - p(p - 1)t^2 + o(t^2)) = -(1/t)/(1 + (p - 1)t/2 + o(t)) = -(1/t)(1 - (p - 1)t/2 + o(t)) = -1/t + (p - 1)/2 + o(1), t -> 0
Same for the term with q. So:
p/(1 - (1 + t)^p) - q/(1 - (1 + t)^q) = -1/t + (p - 1)/2 + o(1) - (-1/t + (q - 1)/2 + o(1)) = (p - q)/2 + o(1) -> (p - q)/2, t -> 0

[midnight jay]

I used ur idea in first step only but I substituted x as 1/t. Here is the solution I got

[midnight jay]

,rotate

[midnight jay]

@harsh plume

Please check

[harsh plume]

Oh, smart idea!

[midnight jay]

But the substitution was ur idea to begin with

I just modified it

[harsh plume]

Well, your substitution is different, as is the idea. So, good job!

[midnight jay]

As I don't know college level expansions, I had to come up with something else

Thanks sir! Couldn't be possible without ur help

[harsh plume]

Such a thing (expressing something in terms of itself) sometimes comes up in integrals, but I don't think I've ever seen it come up in limits. Cool!

[harsh plume]

You're welcome!

[midnight jay]

+close