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Topic: Sequence limit
Replies: 72   Last Post: Nov 26, 2013 12:07 AM

 Messages: [ Previous | Next ]
 Robin Chapman Posts: 412 Registered: 5/29/08
Re: Sequence limit
Posted: Oct 7, 2013 1:46 PM

On 07/10/2013 15:34, Mohan Pawar wrote:

> I won't attempt to attain a low level to match you in ranting and
> displaying vitriol. At times it is better to keep shut and let others
> wonder that you are stupid than to open your mouth and remove their all
> doubts. Bart Goddard, when others critically read the solution under
> "somehow explained to Bart Goddard" you will have removed their doubts

Bart might be a bit blunt at times, but he is one of the most
knowledgable contributors remaining in sci.math.

> Let y = |sin (1/m) |^(m) Taking log on both sides => ln(y) = m
> ln(|sin (1/m)|) Take limit on both sides as m->0 and evaluating it
> =>lim m->0 ln(y) = lim m->0 m ln(|sin (1/m)|)= 0 (at the time of
> evaluating limit, m=0 is the multiplier and one doesn?t need to care
> about value of ln(|sin (1/m)|). Also, the limit is determinate.)

You do need to care about the value of log |sin(1/m)|.
Note that here m = 1/n in the original problem, so that
m is a reciprocal of a natural number. With this class of m,
1/m can be arbitrarily close to an integer multiple of pi, and
so |sin(1/m)| can be arbitarily close to zero and log|sin(1/m)|
can be the negative of an arbitrarily large number. The question
is, whether then multiplying log|sin(1/m)| will give you a sequence
tending to zero or not. A priori, |log|sin(1/m)|| might get large
enough so that even m|log|sin(1/m)|| doesn't tend to zero.
(It doesn't but that needs some nontrivial mathematics).

You should realize that if you have two sequences (a_n) converging
to zero and (b_n) then it is not automatic that the sequence
(a_n b_n) converges to zero. To determine it limiting behaviour,
(Your botched argument is the case a_n = 1/n and b_n = log|sin(1/n)|;
certainly a_n -> 0 but that is not enough by itself to conlcude
a_n b_n -> 0. Note that (b_n) is here an *unbounded* sequence.)

> =>lim m->0 ln(y) = 0 => lim m->0 y = e^0=1 => lim m->0 |sin (1/m)
> |^(m)= 1 as before. (ALSO, VERIFIABLE ON WOLFRAM ALPHA)

Well, if Wolfie Alpha says it's true, it must be true :-)

Returning to the original and less confusing notation then
(i) the sequence (c_n) defined for natural numbers n by
c_n = |sin(n)|^{1/n} converges to 1, and
(ii) the function f(x) defined for real x>0 by
f(x) = |sin(x)|^{1/x} does not have a limit as x -> infinity.

As we have seen, (ii) is easy, but (i) is relatively difficult
to prove.

Date Subject Author
10/3/13 Bart Goddard
10/3/13 Karl-Olav Nyberg
10/3/13 quasi
10/3/13 quasi
10/3/13 Karl-Olav Nyberg
10/3/13 quasi
10/4/13 Roland Franzius
10/4/13 quasi
10/5/13 Roland Franzius
10/5/13 quasi
10/26/13 Roland Franzius
10/26/13 karl
10/26/13 Roland Franzius
10/26/13 gnasher729
10/27/13 karl
10/3/13 quasi
10/4/13 Leon Aigret
10/4/13 William Elliot
10/4/13 quasi
10/4/13 William Elliot
10/4/13 quasi
10/4/13 David C. Ullrich
10/4/13 Robin Chapman
10/5/13 Bart Goddard
10/4/13 Bart Goddard
10/4/13 Peter Percival
10/5/13 Virgil
10/4/13 Bart Goddard
10/6/13 David Bernier
10/6/13 Virgil
10/6/13 Bart Goddard
10/7/13 Mohan Pawar
10/7/13 Bart Goddard
10/7/13 gnasher729
10/7/13 Richard Tobin
10/7/13 Robin Chapman
10/7/13 Michael F. Stemper
10/7/13 Michael F. Stemper
10/7/13 David Bernier
10/7/13 fom
10/8/13 Virgil
10/8/13 fom
10/8/13 Virgil
10/8/13 fom
10/4/13 fom
10/4/13 quasi
10/4/13 quasi
10/9/13 Shmuel (Seymour J.) Metz
10/10/13 Bart Goddard
11/5/13 Shmuel (Seymour J.) Metz
11/6/13 Bart Goddard
11/11/13 Shmuel (Seymour J.) Metz
11/12/13 Bart Goddard
11/15/13 Shmuel (Seymour J.) Metz
11/15/13 Bart Goddard
11/6/13 Timothy Murphy
11/8/13 Bart Goddard
11/8/13 Paul
11/8/13 Bart Goddard
11/9/13 Paul
11/9/13 quasi
11/9/13 quasi
11/9/13 quasi
11/13/13 Timothy Murphy
11/13/13 quasi
11/14/13 Timothy Murphy
11/14/13 Virgil
11/14/13 Roland Franzius
11/26/13 Shmuel (Seymour J.) Metz
11/9/13 Roland Franzius
11/9/13 Paul