evaluation of the limit given by $\lim_{n\to \infty} \sin( (2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$

evaluate :
$\lim_{n\to \infty} \sin( (2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$
I am not sure but I think the answer will be $0$. Can anybody verify me please.

-
Can you please elaborate why you think the answer is zero? What did you do so far? Here is a big hint: if you do the "obvious" thing the limit becomes easy... – N. S. Jan 19 '13 at 17:06
I am not sure but guess it may be zero because $\sin (2n \pi)$ is zero – sumon Jan 19 '13 at 17:10

Obviously, $$\sin[( 2n\pi +\frac 1{2n\pi}) \sin(2n\pi +\frac 1{2n\pi})]=\sin[( 2n\pi +\frac 1{2n\pi}) \sin\frac 1{2n\pi}]$$ Observe that $$\lim_{n\to \infty}2n\pi \sin\frac 1{2n\pi}=\lim_{x\to \infty}x\sin\frac 1x=\lim_{h\to 0^+}\frac 1h\sin h=1$$ and $\frac 1{2n\pi}\sin\frac 1{2n\pi}\to 0$.

Therefore, $$\sin[( 2n\pi +\frac 1{2n\pi}) \sin\frac 1{2n\pi}]\to \sin1$$

-
+1 Very nice and well-explained. – DonAntonio Jan 19 '13 at 17:34

Recall that :

‎$‎sin(2‎n\pi+‎\alpha‎‎)‎=sin ‎\alpha‎‎‎$‎‎ , ‎$‎\lim_{‎n‎\rightarrow \infty‎‎}‎‎\frac{1}{‎2n\pi‎}‎sin(‎\frac{1}{‎2n\pi‎}‎)=0‎$‎ , $‎\lim_{‎n‎\rightarrow \infty‎‎}‎‎2n\pi‎sin(‎\frac{1}{‎2n\pi‎}‎)=1‎$

Therefore

L= $‎\lim_{‎n‎\rightarrow \infty‎‎}sin(‎‎2n\pi‎sin(‎\frac{1}{‎2n\pi‎}‎)+‎‎\frac{1}{‎2n\pi‎}‎sin(‎\frac{1}{‎2n\pi‎}‎))=sin 1^{\circ}$

-