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  • How can I evaluate   <img alt="\lim _{n \rightarrow \infty} \frac{n \sin n}{2 n^2-1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bn%20%5Crightarrow%20%5Cinfty%7D%20%

How can I evaluate   \lim _{n \rightarrow \infty} \frac{n \sin n}{2 n^2-1}

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

Since  -1 \leq \sin x \leq 1  for all x , we have

                                \left|\frac{n \sin n}{2 n^2-1}\right|=\frac{n|\sin n|}{2 n^2-1} \leq \frac{n}{2 n^2-1} ;

and since 2n^2-1 \geq n^2\: \: for\: \: all\: \: n \geq 1,
                                \left|\frac{n \sin n}{2 n^2-1}\right| \leq \frac{n}{n^2}=\frac{1}{n} \underset{n \rightarrow \infty}{\longrightarrow} 0

Posted by

Gunjita

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