I'm not understanding this limit problem...

lim of (n)sin(1/n) as n goes to infinity. I thought the limit would be 0 since sin[1/(really big number)] = 0, but the book says the limit is 1. how do you get the limit as 1? Thanks.

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Originally Posted by virtuoso735

lim of (n)sin(1/n) as n goes to infinity. I thought the limit would be 0 since sin[1/(really big number)] = 0, but the book says the limit is 1. how do you get the limit as 1? Thanks.

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Let . Then as , . Thus,

Quote:

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Originally Posted by virtuoso735

lim of (n)sin(1/n) as n goes to infinity. I thought the limit would be 0 since sin[1/(really big number)] = 0, but the book says the limit is 1. how do you get the limit as 1? Thanks.

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Dont forget this
Its not just

sin[1/(really big number)]

But its

(really big number) x (sin[1/(really big number)])

So can't just do it that way

However put 1/n =t

So when n-> infinity , t-> 0
Hence it becomes

Lt_{t->0} (sin[t]/t) = 1

For the geometrical proof of this
Read this
--thanks to Chris for it

EDIT: Chris won!! :D