How do I evaluate this limit that seems to be indeterminate?

Question

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?

Answer 1

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 $x \to 0$

$$\frac{\sin x-x}{x^3}\to-\frac 16 $$

Answer 2

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...