```Date: Oct 4, 2013 4:42 PM
Author: quasi
Subject: Re: Sequence limit

quasi wrote:>Mohan Pawar wrote:>>>>Let>>>>    x=1/m where m is real, -inf< x and m <inf. >>>>=>     as x->inf., m->0 >>=>     lim x -> inf. |sin x|^(1/x)>>= lim m -> 0 |sin (1/m) |^(m)>>= 1  (see below why 1)>>No, it's not equal to 1. >>In fact, The limit>>   lim (m --> 0) |sin(1/m)|^(1/m)I meant: lim (m --> 0) |sin(1/m)|^m>does not exist.>>>Note that the value of |sin(1/m)| varies from 0 to to 1>>BUT exponent m is guaranteed to be zero as m -> 0.>>No. The exponent m is only guaranteed to _approach_ 0.>>As m approaches 0, there are infinitely many values of m such>that sin(1/m) = 0. For those values of m,>>   |sin(1/m)|^(1/m) = 0 I meant: |sin(1/m)|^m = 0 >hence for those values of m,>>   |sin(1/m)|^(1/m)I meant: |sin(1/m)|^m>approaches 0. >>On the other hand, as m approaches 0, there are infinitely>many values of m such that sin(1/m) = 1. For those values of m,>>   |sin(1/m)|^(1/m) = 1I meant: |sin(1/m)|^m = 1>hence for those values of m, >>   |sin(1/m)|^(1/m)I meant: |sin(1/m)|^m>approaches 1.>>It follows that the limit>>   lim (m --> 0) |sin(1/m)|^(1/m)I meant: lim (m --> 0) |sin(1/m)|^m>does not exist.>>In fact, for any real constant c between 0 and 1 inclusive,>there exists an infinite sequence of values of m approaching>zero such that |sin(1/m)|^(1/m) = c.I meant: such that |sin(1/m)|^m = c.Ugh.By obliviously copying and pasting one typo many times, I endedup with a lot of typos.Sorry for the confusion.quasi
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