
Re: Sequence limit
Posted:
Oct 7, 2013 11:43 AM


Mohan Pawar <mohanpawar@mpclasses.com> wrote in news:31dc2367384641cfbd3cea1f796ba7c8@googlegroups.com:
> I won't attempt to attain a low level to match you
That "low level" is still far above you.
> 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).
No, the relevant issue is that there is a difference between the discrete and continuous versions of the limit. You remain completely ignorant of the whole issue.
You also continue to ignore the fact that sin(x) is zero infinitely often, and therefore is very close to zero infinitely often. Your garbage proof skips over this most important fact when you say
> and one doesn't need to care > about value of ln(sin (1/m)).
Sure, skip the most important fact without justification.
> (ALSO, VERIFIABLE ON WOLFRAM ALPHA)
Which assumes a continuous variable, and therefore is not relevant to the issue here.
> Notice that the new additional steps are no different from my > original two line justification that saves above labor
Yes, both versions are equally incorrect.
Please go away. Your unfounded arrogance is far worse than what you call my "vitriol."
If you ever get to the point where you understand that
lim_{x> oo} (sin(x))^{1/x} does not exist, while your "proof" above shows the opposite, you'll be at college freshman level. You still won't have anything interesting to say, but at least you'll have made progress.

