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Evaluate

$$\lim_{x\to 1} \frac{p}{1-x^p}-\frac{q}{1-x^q}$$ where $p,q$ are Natural Numbers.

I tried rationalization, but I wasn't able to get anywhere. I'm not being able to remove the $\infty-\infty$ indeterminate form. Any help would be appreciated. Many thanks!

EDIT: The given answer is $\frac{p-q}{2}$

Daniel R
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user342209
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  • Are $p,q$ related in any way (besides both being $\in\Bbb{N}$)? – MathematicianByMistake May 31 '16 at 18:19
  • No Sir. There's no other relation. – user342209 May 31 '16 at 18:21
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    Use the fact that $1-x^n = (1-x)(1+x+x^2+ \dots + x^{n-1})$. Factorize the denominators and then sum the two things. – Crostul May 31 '16 at 18:25
  • Any particular reason why this has to be done without l'Hospital? Doing it twice gives the answer – Wojciech Morawiec May 31 '16 at 18:33
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    @WojciechMorawiec or even just once! – Adam Hughes May 31 '16 at 18:35
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    Actually I'm trying to develop the ability to solve Limits without using L'Hospital because most questions that I'm asked to solve cannot be solved using L'Hopital. i thus wanted practise of solving Limits without resorting to L'Hopital. – user342209 May 31 '16 at 18:37
  • @user342209 Fair enough, can you clarify though: were you specifically told not to use it or is this just your personal preference? You would probably not be able to think up Jack's trick in a real life situation, and I'm not certain there's a terribly simple direct way of doing this (that a typical student could come up with in real life) without L'Hôpital. – Adam Hughes May 31 '16 at 18:38
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    Personal preference predominantly, but my instructor also urges us not to use L'Hopital. Actually I'm preparing for the JEE exam here in India and questions on Limits become nearly impossible to solve using L'Hopital. – user342209 May 31 '16 at 18:39
  • @user342209 Good luck to you then. A piece of professional advice though: never limit yourself on what tools you can use: if no L'Hôpital turns out to be the best for a given problem, don't use it, but if L'Hôpital is, do use it, ultimately what's important is you know what the right tool is for the right job, otherwise you'll end up making your life harder for no good reason. It sounds like you need practice with non L'Hôpital techniques and practice in recognizing when the rule will or will not help you (in this case both ways are possible). – Adam Hughes May 31 '16 at 18:43
  • Thanks Sir. I'll definitely keep your advice in mind.And you're right Sir, I do need practise with both. Thanks once again Sir. – user342209 May 31 '16 at 19:30
  • @AdamHughes Could you please suggest a book from where I could learn methods of solving questions on Limits, Differentiation, Integration and Calculus as a whole? As such I used to study from Apostol's books on Calculus, but they aren't of much help in learning problem solving techniques. – user342209 May 31 '16 at 19:32
  • @user342209 I think the gold standard is usually considered to be Spivak's book. I would definitely start there if you're serious about learning calculus. – Adam Hughes May 31 '16 at 19:35
  • Thanks very much Sir! I'll certainly go through the book. Actually Calculus is my favourite part of Maths as a whole! – user342209 May 31 '16 at 19:43
  • Related http://math.stackexchange.com/questions/1796855/find-the-limit-lim-x-to-1-left-fracp1-xp-fracq1-xq-right/1796889#1796889 – cgiovanardi May 31 '16 at 22:33
  • @user342209 Apologies for disturbing this comment thread, but you said you were preparing for JEE. I presume JEE Advanced 2017. How were your results? Did you get into any IITs? – Bongo Man Jan 03 '24 at 13:46

2 Answers2

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Symmetry! Let $f_{p,q}(x) = \frac{p}{1-x^p}-\frac{q}{1-x^q}$. Then: $$ f_{p,q}\left(x^{-1}\right) = (p-q)-f_{p,q}(x)$$ hence, assuming that the original limit exists:

$$ \lim_{x\to 1}f_{p,q}(x) = \frac{1}{2}\lim_{x\to 1}\left(f_{p,q}(x)+f_{p,q}(x^{-1})\right) = \frac{1}{2}\lim_{x\to 1}(p-q)=\color{red}{\frac{p-q}{2}}.$$

The assumption $p,q\in\mathbb{N}^+$ is in fact unnecessary, we just need $p,q\neq 0$.

Jack D'Aurizio
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I think that there is a flaw in Jack's proof. I find that approach interesting but it's kind of an ansatz, because it assumes that the limit exists. That method is useful to prove that the limit must be $\frac{p-q}{2}$ if it exists. When we do not know whether the limit exists or not, we may come to a wrong conclusions if we don't take care. I'll give you an example using additive symmetry instead of multiplicative symmetry.

Consider $f:\mathbb{R}-\{0\}\to \mathbb{R}$ given by $f(x) = \sin(\frac{1}{x})$. It is clear that $f(x) + f(-x) = 0$ so we have $$ \lim_{x\to 0}f(x) = \frac{1}{2}\lim_{x\to 0}\left(f(x)+f(-x)\right) = \frac{1}{2}\lim_{x\to 0}0=0$$

But that's wrong as the limit $\lim_{x\to 0}f(x)$ does not exist.

To overcome this difficulty you have two options. The first one is to complement Jack's argument with a proof of existence of the limit. Another possibility is to provide a completely different proof. For the first case, I guess that it should be not too complicated to prove that $f_{p,q}$ is monotone on an interval $(1-\epsilon;1)$. For the second, I'll give another proof working OK for the case $p,q \in \mathbb{N}$.

Define $f_{p}(x) = \frac{1-x}{1-x^p} = \frac{1}{1+x+ \cdots + x^{p-1}}$ for $p \in \mathbb{N}$. With the second expression we can think that $f_{p}$ is defined over $\mathbb{R}$. Moreover, we have $f_p \in C^{\infty}(\mathbb{R})$. Then, computing the first derivative and using differentiability of $f_p$ at $1$ we have $$f_p(x) = \frac{1}{p} - \frac{p-1}{2p}(x-1) + \varepsilon(x-1)$$ where $\frac{\varepsilon(x)}{x} \to 0$ as $x$ approaches $0$. Now, for $x\neq1$ you have $$ \frac{p}{1-x^p} = \frac{p}{1-x} \cdot f_p(x) = \frac{1}{1-x} + \frac{p-1}{2} - p\frac{\varepsilon(x-1)}{x-1}$$

Considering $f_p$ and $f_q$ and doubling the previous manipulations, we have $$ \frac{p}{1-x^p} - \frac{q}{1-x^q} = \frac{p-q}{2} + \delta (x)$$ with $\delta (x) \to 0$ when $x$ approaches $1$ because of the aforementioned property for $\varepsilon$.

Pipicito
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