- #1
Buffu
- 849
- 146
$$x_n = (-1)^n {2n\over n+1} \sin n $$
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
$$\lim_{n \to \infty} |(-1)^n| {2n\over n+1} |\sin n|$$
$$\lim_{n \to \infty} {|(-1)^n| \over n} {2n^3\over n+1} {|\sin n| \over n}$$
$$\lim_{n \to \infty} {|(-1)^n| \over n}{2\over {1\over n^2}+{1\over n^3}} {|\sin n |\over n} ={2 \over 0}$$
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
$$\lim_{n \to \infty} |(-1)^n| {2n\over n+1} |\sin n|$$
$$\lim_{n \to \infty} {|(-1)^n| \over n} {2n^3\over n+1} {|\sin n| \over n}$$
$$\lim_{n \to \infty} {|(-1)^n| \over n}{2\over {1\over n^2}+{1\over n^3}} {|\sin n |\over n} ={2 \over 0}$$