<s>[INST] 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?
 [/INST] 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 $x \to 0$
$$\frac{\sin x-x}{x^3}\to-\frac 16 $$
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