What to think about this limit trick $\lim_{x\to 1}f(x)=\lim_{x\to 1}f(\frac 1x)$? While searching for duplicates to a question asked recently I stumbled onto this specific answer:
Show that $\lim\limits_{x\to 1}\left(\frac p{1-x^p}-\frac q{1-x^q}\right)=\frac {p-q}2$
I reproduce the sketch here:

$$L=\lim\limits_{x\to 1}\left(\frac p{1-x^p}-\frac q{1-x^q}\right)=\lim\limits_{x\to 1}\left(\frac p{1-(\frac 1x)^p}-\frac q{1-(\frac 1x)^q}\right)=\lim\limits_{x\to 1}\left(\frac {-px^p}{1-x^p}-\frac {-qx^q}{1-x^q}\right)$$
So by adding first and third term we get $2L=p-q$.

What annoys me is that it looks like an answer we could put in the "bad math that gets away with it" thread...
It is true that $\lim\limits_{x\to 1}f(x)=\lim\limits_{x\to 1}f(\frac 1x)$ when the limit exists, however here, I think it is part of the problem to show that this limit actually exists.
Looking at it more closely we have that:

*

*$f_p(x)=\dfrac{p}{1-x^p}=\frac 1{1-x}+\frac{n-1}2+O(1-x)$

*$f_p(\frac 1x)=\dfrac{p}{1-(\frac 1x)^p}=-\frac 1{1-x}+\frac{n+1}2+O(1-x)$
So if we were to apply the same process to $\ell_p=\lim\limits_{x\to 1}\dfrac{p}{1-x^p}$ we would get $2\ell_p=p\iff \ell_p=\frac p2$ which is wrong since $1$ is a pole here (limit does not exist).
But the solution proposed in the cited answer let us think that we got the result from $\ell_p-\ell _q=\frac p2-\frac q2=\frac{p-q}2$
While in reality we get it from cancellation of the divergent parts in:
$\require{cancel}f_p(x)-f_q(x)=\left(\cancel{\frac 1{1-x}}+\frac 12(p-1)+O(1-x)\right)-\left(\cancel{\frac 1{1-x}}+\frac 12(q-1)+O(1-x)\right)=\frac{(p-\cancel{1})-(q-\cancel{1})}2+O(1-x)\to \frac{p-q}2$
So what is your opinion on this answer, do you agree with my point of view ?
How would you convince a student that does this, that his method is flawed, or at least that he should prove first that the limit exists before carrying the $\lim$ operator from equality to equality like this (which BTW seems to be a very common habit among posters...).
 A: I wouldn't quite call this bad math, sometimes you just have to read from right to left.
Often people write $\lim_{x\to a} f(x) = \lim_{x \to a} g(x)$ to mean that the limits either both exist or both don't exists, and eventually continue the chain showing the right one exists, meaning they both exist.
Recall that if $\lim_{x \to a} g(y) = L$ and $\lim_{y \to L}f(y) = M$, then $\lim_{x \to a} f(g(x)) = M$. Apply to $g(x) = 1/x$ at $x=1$. We see that if $\lim_{y \to 1} f(y)$ exists and equals $M$, then $\lim_{x \to 1} f(1/x)$ exists and equals $M$. Now apply this again to  $h(x) = f(1/x)$ to see that if $\lim_{y \to 1}f(1/y)$ exists and equals $M$, then $\lim_{x \to 1} f(1/(1/x)) = \lim_{x \to 1}f(x)$ exists and equals $M$.
Thus we see that $\lim_{x \to 1} f(x)$ exists and equals $M$ if and only if $\lim_{x \to 1} f(1/x)$ exists and equals $M$. Thus it is not really "bad math" to write $\lim_{x \to 1} f(x) = \lim_{x \to 1} f(1/x)$ as long as you eventually show that one of the two sides exists.
