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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- Mar 29th 2009, 09:17 PMvirtuoso735I'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.

- Mar 29th 2009, 09:32 PMChris L T521
- Mar 29th 2009, 09:52 PMADARSH
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**--thanks to Chris for it

For the geometrical proof of this

Read this

EDIT: Chris won!! :D