fom
Posts:
1,968
Registered:
12/4/12


Re: Sequence limit
Posted:
Oct 7, 2013 11:52 PM


On 10/7/2013 10:33 PM, David Bernier wrote: > 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... >
Having fun with this one, are you?
:)

