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) = 1


I 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 ended
up with a lot of typos.

Sorry for the confusion.

quasi