
Re: Sequence limit
Posted:
Oct 7, 2013 11:33 PM


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." >
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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'uh willing.

