How do I evaluate this limit that seems to be indeterminate? I've been trying to evaluate the below limit, which Mathematica claims is equals to $1$.
$$\underset{n\to \infty }{\text{lim}}\frac{\frac{1}{2} (n+1) \sin \left(\frac{2 \pi }{n+1}\right)-\pi }{\frac{1}{2} n \sin \left(\frac{2 \pi }{n}\right)-\pi }$$
However, my attempts, of which I attempted using L'Hopital's rule, always end with the indeterminate form of $\frac{0}{0}$, because taking the limit to infinity of both the non-$\pi$ sections of the numerator and denominators gives $\pi$.
Is Mathematica correct, and if so, what is the solution?
 A: HINT
We have that
$$\frac{\frac{1}{2} (n+1) \sin \left(\frac{2 \pi }{n+1}\right)-\pi }{\frac{1}{2} n \sin \left(\frac{2 \pi }{n}\right)-\pi }=
\frac{{\sin \left(\frac{2 \pi }{n+1}\right)}-\frac{2 \pi }{n+1} }{\left(\frac{2 \pi }{n+1}\right)^3}
\frac{\left(\frac{2 \pi }{n} \right)^3}{{\sin \left(\frac{2 \pi }{n}\right)}-\frac{2 \pi }{n} }
\frac {n^2} {(n+1)^2}$$
then it suffices to show that as $x \to 0$
$$\frac{\sin x-x}{x^3}\to-\frac 16 $$
A: I'm a friend of the OP irl, and I suggested the following attempt at a (possibly bogus) proof. (I know I should probably comment this, but I don't have enough reputation to do so -_-) Hopefully someone more qualified can check this?
Consider the general case:
$$ \lim_{x -> \infty}\frac{\frac{\left(x+k\right)}{2}\sin\left(\frac{2\pi}{x+k}\right)-\pi}{\frac{\left(x\right)}{2}\sin\left(\frac{2\pi}{x}\right)-\pi} $$
We will attempt strong induction on $k$.
The base case $k=0$ is:
$$ \lim_{x -> \infty}\frac{\frac{\left(x\right)}{2}\sin\left(\frac{2\pi}{x}\right)-\pi}{\frac{\left(x\right)}{2}\sin\left(\frac{2\pi}{x}\right)-\pi} $$
which is trivially equal to 1.
Then, the induction hypothesis is:
$\forall m \geq n, n \in \mathbb{Z^{+}}$:
$$ \lim_{x -> \infty}\frac{\frac{\left(x+m\right)}{2}\sin\left(\frac{2\pi}{x+m}\right)-\pi}{\frac{\left(x\right)}{2}\sin\left(\frac{2\pi}{x}\right)-\pi} = 1$$
Here, we should be able to substitute $x$ for something like $t =x-1$. From the induction hypothesis:
$$ \lim_{t -> \infty}\frac{\frac{\left(t+m+1\right)}{2}\sin\left(\frac{2\pi}{t+m+1}\right)-\pi}{\frac{\left(t+1\right)}{2}\sin\left(\frac{2\pi}{t+1}\right)-\pi} = 1$$
Also from the induction hypothesis for $n > 0$:
$$ \lim_{t -> \infty}\frac{\frac{\left(t+1\right)}{2}\sin\left(\frac{2\pi}{t+1}\right)-\pi}{\frac{\left(t\right)}{2}\sin\left(\frac{2\pi}{t}\right)-\pi} = 1$$
So we can use product rule here and say:
$$\lim_{t -> \infty}(\frac{\frac{\left(t+1\right)}{2}\sin\left(\frac{2\pi}{t+1}\right)-\pi}{\frac{\left(t\right)}{2}\sin\left(\frac{2\pi}{t}\right)-\pi})(\frac{\frac{\left(t+m+1\right)}{2}\sin\left(\frac{2\pi}{t+m+1}\right)-\pi}{\frac{\left(t+1\right)}{2}\sin\left(\frac{2\pi}{t+1}\right)-\pi}) = 1 $$
$$ \lim_{t -> \infty}\frac{\frac{\left(t+m+1\right)}{2}\sin\left(\frac{2\pi}{t+m+1}\right)-\pi}{\frac{\left(t\right)}{2}\sin\left(\frac{2\pi}{t}\right)-\pi} = 1 $$
I'm a bit concerned about this step here, but I don't see why I can't change $t$ to $x$ even though the initial substitution was $t=x-1$. If this is not...illegal...then this should demonstrate the induction step, and show that $\forall k \in \mathbb{Z^{+}}$:
$$ \lim_{x -> \infty}\frac{\frac{\left(x+k\right)}{2}\sin\left(\frac{2\pi}{x+k}\right)-\pi}{\frac{\left(x\right)}{2}\sin\left(\frac{2\pi}{x}\right)-\pi}  = 1$$
And thus when $k=1$, $$ \lim_{x -> \infty}\frac{\frac{\left(x+1\right)}{2}\sin\left(\frac{2\pi}{x+1}\right)-\pi}{\frac{\left(x\right)}{2}\sin\left(\frac{2\pi}{x}\right)-\pi} = 1$$ too right?
Checking with WolframAlpha, the general case doesn't seem to be FALSE per se, but this "proof" just feels very dubious to me...
