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The Limit of Sin(1/x)


Date: 10/02/98 at 16:21:10
From: Bryan Johnson
Subject: lim of sin (1/x)

My math teacher gave us this problem to do:
  
   lim  sin (1/x)
   x -> 0

My reasoning is as follows. For the limit to exist, the limit must be 
the same approaching both sides of zero. 

For:
 
   lim   sin (1/x)  
   x -> 0+               

as x gets really tiny from the positive side, (1/x) gets really big 
(positive), and looking at the y = sin x curve, as x gets really big 
on the positive side, there is no limit.  
 
For:

   lim   sin (1/x)   
   x -> 0-              

as x gets really small from the negative side, (1/x) gets really big 
(negative), and looking at the y = sin x curve, as x gets really big 
on the negative side, there is no limit.

Since both limits appear to be DNE (does not exist), then would:

   lim  sin (1/x)   
   x -> 0                         


be DNE also?

Thanks for your time,
Bryan


Date: 10/02/98 at 17:03:45
From: Doctor Anthony
Subject: Re: lim of sin (1/x)

As x approaches 0, 1/x increases without bound. Thus in any open 
interval containing 0 there will be values of x such that 1/x is a 
multiple of 2 pi, values of x such that 1/x is pi/2 more than a 
multiple of 2 pi, and values of x such that 1/x is (3 pi)/2 more than 
a multiple of 2 pi. The sine of these values of x will be 0, 1 and -1 
respectively.

Sin(1/x) oscillates "an infinite number of times" between 1 and -1
in any neighbourhood of x -> 0. Therefore the limit doesn't exist.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Analysis
High School Calculus

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