# 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... :)

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I am not seeing here any mistake – Dutta Dec 14 '13 at 14:20
This is in NBHM 2013 Ph.D exam question paper... – Praphulla Koushik Dec 14 '13 at 14:21
I did not see your heading first – Dutta Dec 14 '13 at 14:22
It is alright :) I have kept it just for my quick reference.... :) – Praphulla Koushik Dec 14 '13 at 14:23
@PraphullaKoushik: really, this is a Ph.D. exam question? Seems pretty basic for that. – Eckhard Dec 14 '13 at 14:23

## 1 Answer

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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