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Is there a possible way to compute this limit ? I tried the L'Hôpital's rule, but it seems like the function become more complicated.

$$\lim_{n\to\infty} \frac{1}{n \log(n)} \frac{\sin\left(\frac{\pi}{n}+\frac{2 \pi \log(n)}{\sqrt n}\right)}{\sin\left(\frac{\pi}{n}\right)}$$

All suggestions are welcomed.

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First, $$\lim_n\frac{\sin\frac\pi n}{\frac1n}=\pi,$$ so you might as well calculate $$ \lim_{n\to\infty} \dfrac{\sin(\frac{\pi}{n}+\frac{2 \pi \log(n)}{\sqrt n})}{\log n}. $$ You have $$ \left|\dfrac{\sin(\frac{\pi}{n}+\frac{2 \pi \log(n)}{\sqrt n})}{\log n}\right|\leq\frac1{\log n}\xrightarrow{\ \ \ \ \ }0, $$ so the limit is zero.

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  • $\begingroup$ You misses something on the $ n \to $ in the first term. $\endgroup$ – Royi May 10 at 7:57
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You can not only find the limit, but find an equivalent when $n\to\infty$:

  • $\sin\biggl(\dfrac\pi n+\dfrac{2 \pi\log n}{\sqrt n}\biggr)\sim_\infty\dfrac\pi n+\dfrac{2 \pi\log n}{\sqrt n}\sim_\infty\dfrac{2 \pi\log n}{\sqrt n}$,
  • $\sin\dfrac\pi n\sim_\infty\dfrac\pi n$ hence $$\frac1{n\log n}\frac{\sin\biggl(\dfrac\pi n+\dfrac{2 \pi\log n}{\sqrt n}\biggr)}{\sin\dfrac\pi n}\sim_\infty\frac1{n\log n}\frac{\dfrac{2 \pi\log n}{\sqrt n}}{\dfrac\pi n}=\color{red}{\frac2{\sqrt n}}\to 0.$$
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Use the inequality $|\sin x|\leq |x|$. We have $$ \begin{align} \left| \dfrac{1}{n \log(n)} \dfrac{\sin(\frac{\pi}{n}+\frac{2 \pi \log(n)}{\sqrt n})}{\sin(\frac{\pi}{n})}\right|\leq \frac{ \frac1{\log n} \left(\frac{\pi}n+\frac{2\pi \log n}{\sqrt n} \right)}{ n \sin(\frac{\pi}n)} \end{align} $$ The numerator approaches $0$ as $n\rightarrow \infty$, but the denominator approaches $\pi$ as $n\rightarrow\infty$.

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$$ A=\dfrac{\sin(\frac{\pi}{n}+\frac{2 \pi \log(n)}{\sqrt n})}{\sin(\frac{\pi}{n})}=\frac{\sin \left(\frac{\pi }{n}+\frac{2 \pi \log (n)}{\sqrt{n}}\right)}{\frac{\pi }{n}+\frac{2 \pi \log (n)}{\sqrt{n}}}\times\dfrac{\frac\pi n}{{\sin(\frac{\pi}{n})}}\times\frac{\frac{\pi }{n}+\frac{2 \pi \log (n)}{\sqrt{n}} }{\frac \pi n}$$ Since $n$ is large, you face twice the limit of $\frac{\sin(x)}x=1$ when $x\to 0$ and so, for large values of $n$ $$A\simeq \frac{\frac{\pi }{n}+\frac{2 \pi \log (n)}{\sqrt{n}} }{\frac \pi n}=2 \sqrt{n} \log (n)+1$$ So, $$\frac A{n\log(n)}\simeq \frac{2 \sqrt{n} \log (n)+1 }{n\log(n)}=\frac{2}{\sqrt n}+\frac 1{n\log(n)}$$ which shows not only the limit but also how it is approached.

Edit

We could even go further using composition of series and show that $$\frac A{n\log(n)}=\frac{2}{\sqrt n}+\frac 1{n\log(n)}-\frac{4 \pi ^2 \log ^2(n)}{3 n^{3/2}}+O\left(\frac{1}{n^2}\right)$$

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