```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...-- Let us all be paranoid. More so than no such agence, Bolon Yokte K'uhwilling.
```