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