
Re: Sequence limit
Posted:
Oct 7, 2013 11:55 AM


On Monday, October 7, 2013 3:34:37 PM UTC+1, Mohan Pawar wrote: > =>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.)
> Mohan Pawar > Online Instructor, Maths/Physics > MP Classes LLC
You are either lying about being a maths instructor (that's the positive possibility), or you are not, in which case you should be removed from your post immediately.
You were calculating the limit of a product f (x) * g (x). In this case, the limit of f (x) is zero. g (x) has no limit, and has no upper bound. As sin (1 / m) comes arbitrarily close to 0 ( and even reaches it), ln sin (1 / m) has no lower bound.
It is quite easy to prove that lim (f (x) * g (x)) is zero if the limit of f (x) is zero and g (x) is bounded. There is nothing that can be said directly if g (x) has no upper and lower bound. You could try to apply l'Hospital's rule, good luck with that.
In summary, you made a beginners mistake. Now making a beginners mistake can happen to anyone. Insisting that you are right is quite damning for your reputation.
On the other hand, the problem is not for real m, but for integers. sin (m) is never 0 for integer m with the exception of m = 0. There are some rather deep results about how well pi can be approximated by rational numbers, and these results might show that the limit sin (m) ^ (1/m) for integer m is indeed 1. For real m, it's not even funny how wrong you are.

