Find the limit $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$$

We have $\sin(2n\pi+\theta) = \sin(\theta)$ this implies $\sin(2n\pi + \frac{1}{2n\pi})) = \sin(\frac{1}{2n\pi}))$ and given limit becomes $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(\frac{1}{2n\pi})).$$

Now, $$\sin((2n\pi + \frac{1}{2n\pi}) \sin(\frac{1}{2n\pi})) = \sin((2n\pi) \sin(\frac{1}{2n\pi})) + \frac{1}{2n\pi}\sin(\frac{1}{2n\pi})) $$

$|sin 2n\pi| \le 1$ and the sequence $\frac{1}{2n\pi}$ converges to zero and hence second term is converges to zero.

$\lim _{\theta \to 0}\frac{sin \theta}{\theta} = 1$ and so $(2n\pi) \sin(\frac{1}{2n\pi})$ converges to 1 as $n \to \infty$. Hence the required limit is $\sin 1$.

**Solution**:We have $\sin(2n\pi+\theta) = \sin(\theta)$ this implies $\sin(2n\pi + \frac{1}{2n\pi})) = \sin(\frac{1}{2n\pi}))$ and given limit becomes $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(\frac{1}{2n\pi})).$$

Now, $$\sin((2n\pi + \frac{1}{2n\pi}) \sin(\frac{1}{2n\pi})) = \sin((2n\pi) \sin(\frac{1}{2n\pi})) + \frac{1}{2n\pi}\sin(\frac{1}{2n\pi})) $$

$|sin 2n\pi| \le 1$ and the sequence $\frac{1}{2n\pi}$ converges to zero and hence second term is converges to zero.

$\lim _{\theta \to 0}\frac{sin \theta}{\theta} = 1$ and so $(2n\pi) \sin(\frac{1}{2n\pi})$ converges to 1 as $n \to \infty$. Hence the required limit is $\sin 1$.

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