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Old 7th November 2011, 04:33 AM   #1 (permalink)
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help!! how to show that "limit (sin n)/n" n-> infinity = 0

as stated in title...

how do i show that
limit (sin n)/n, with n approaching infinity?

if it approaches 0, i understand that i can use L'hopital rule. but with sin x approaching infinity, the values varies? so how do i show the working for it?

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Old 7th November 2011, 07:33 AM   #2 (permalink)
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Re: help!! how to show that "limit (sin n)/n" n-> infinity = 0

limit of sin(n) would be -1<sin(n)<1. that divided by infinity will equal 0 regardless of the sin(n) values.

think you only need to use L'hopital's rule if by replacing the values in your current function, it is still unclear. this one's pretty clear without doing anything.

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Old 7th November 2011, 03:06 PM   #3 (permalink)
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Re: help!! how to show that "limit (sin n)/n" n-> infinity = 0

but my sin value is constantly changing? it will reach both the value, 0 and another value as n approaches infinity?

unlike limit approaching 0, sin 0 = 0, thus i have to use L'hopital rule, getting a cos 0, which is 1.
if i apply L'hopital rule to n apprach infinity, cos infinity have a value that constantly varies? so how can i say that when n appraoching infinity, n = 0?

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Old 7th November 2011, 08:22 PM   #4 (permalink)
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Re: help!! how to show that "limit (sin n)/n" n-> infinity = 0

P.S:
nvm i get it already... asking this question coz i had the impression that sin would reach sin 180, etc....

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Old 10th November 2011, 07:31 PM   #5 (permalink)
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Re: help!! how to show that "limit (sin n)/n" n-> infinity = 0

We have 3 common approach to such problem:

1. Squeeze Theorem
2. Taylor approximation (or small angle approximation)
3. L'Ho Rule

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