<s>[INST] Evaluate limit of a sequence... NBHM $2013$ Question is to Evaluate $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$$
All I could do was to see that $$\sin(2n\pi + \frac{1}{2n\pi}))=\sin( \frac{1}{2n\pi})$$
Just because $\sin(2n\pi+\theta)=\sin(\theta)$..
So, we now have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin( \frac{1}{2n\pi}))$$
Now, as $\lim _{x\rightarrow \infty}x\sin(\frac{1}{x})=1$ we would have 
$$\lim_{n\rightarrow \infty}2n\pi \sin( \frac{1}{2n\pi})=1$$
So, we now have $$\lim_{n\rightarrow \infty} \sin(1 + \frac{1}{2n\pi} \sin( \frac{1}{2n\pi}))$$
Now, as $\sin(x)$ is bounded and $\frac{1}{2n\pi} \rightarrow 0$ we would have 
$$\lim_{n\rightarrow \infty}\frac{1}{2n\pi} \sin( \frac{1}{2n\pi})=0$$
So, we would now left with :
$$\lim_{n\rightarrow \infty} \sin(1)=\sin(1)$$
After all i would like to say that as $\sin (x)$ is continuous I can take limits inside.
So, we have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))=\sin 1$$
I would like somebody to check if I have done correctly and I would be thankful if any body can let me know if there is anything more to be specified to do so.
Thank you... :)
 [/INST] Question is to Evaluate $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$$
As $\sin(2n\pi+\theta)=\sin(\theta)$ we would have :
$$\sin(2n\pi + \frac{1}{2n\pi}))=\sin( \frac{1}{2n\pi})$$
So, we now have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin( \frac{1}{2n\pi}))$$
Now, as $\lim _{x\rightarrow \infty}x\sin(\frac{1}{x})=1$ we would have 
$$\lim_{n\rightarrow \infty}2n\pi \sin( \frac{1}{2n\pi})=1$$
So, we now have $$\lim_{n\rightarrow \infty} \sin(1 + \frac{1}{2n\pi} \sin( \frac{1}{2n\pi}))$$
Now, as $\sin(x)$ is bounded and $\frac{1}{2n\pi} \rightarrow 0$ we would have 
$$\lim_{n\rightarrow \infty}\frac{1}{2n\pi} \sin( \frac{1}{2n\pi})=0$$
So, we would now left with :
$$\lim_{n\rightarrow \infty} \sin(1)=\sin(1)$$
After all i would like to say that as $\sin (x)$ is continuous I can take limits inside.
So, we have $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))=\sin 1$$
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