Evaluate $\lim_{x\to 1} \frac{p}{1-x^p}-\frac{q}{1-x^q}$

Solution 1:

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 $p,q\neq 0$.

Solution 2:

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 $\varepsilon$.