$$x_n = (-1)^n {2n\over n+1} \sin n $$(adsbygoogle = window.adsbygoogle || []).push({});

it is given that ,

$$|x_n| = |(-1)^n| {2n\over n+1} |\sin n| < {2n\over n+1} < 2$$

thus bounded, but what i did not get is how did we find ##\lim_{n \to \infty} |(-1)^n| {2n\over n+1} |\sin n| = 2##.

I checked with wolfram alpha and it says ##\lim_{n \to \infty} |(-1)^n| {2n\over n+1} |\sin n| = 2##.

https://www.wolframalpha.com/input/?i=lim+n+to+infinity+(-1)^n+*+sin+n+*+(2n/(n+1))

My try

$$\lim_{n \to \infty} |(-1)^n| {2n\over n+1} |\sin n|$$

$$\lim_{n \to \infty} {|(-1)^n| \over n} {2n^3\over n+1} {|\sin n| \over n}$$

$$\lim_{n \to \infty} {|(-1)^n| \over n}{2\over {1\over n^2}+{1\over n^3}} {|\sin n |\over n} ={2 \over 0}$$

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# I Why this limit exists?

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