Date: Oct 7, 2013 11:33 PM
Author: David Bernier
Subject: Re: Sequence limit

On 10/07/2013 10:34 AM, Mohan Pawar scribbled:
> *********************************************************
> "Somehow" explained to Bart Goddard
> *********************************************************
> I am assuming that the most relevant issue was why limit was decided by the index m and not the base |sin(1/m)| in lim m -> 0 |sin (1/m) |^(m).
> The original problem was transformed into equivalent problem as below:
> Find lim m -> 0 |sin (1/m) |^(m)
> 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.)
> =>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)
> Notice that the new additional steps are no different from my original two line justification that saves above labor except now the index m is brought down as multiplier through log operation. It is still the exponent m now as multiplier that _alone_ decided value of limit. For reference, the original two line justification from my previous post is quoted below:
> "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.
> Now if m is replaced by natural number n, the situation does not change |sin (1/n)|will still be within 0 to 1 and limit will evaluate to due to zero in exponent."


Everybody knows that your proof is leaking,
Everybody knows that it's gonna sink,
That's how it goes,
Everybody knows...

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