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/ |
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