Interesting double Limit

Question

While computing The Fourier transform of a function tending to become a simple blip I came across $$\underset{w\to 0}{\lim_{T\to 0}} \left[ \frac{\sin^2\left(\frac{wT}{2}\right)}{\omega^2 T}\right] $$

I think we can split this into $$ \underset{w\to 0}{\lim_{T\to 0}}\left[\frac{\sin\left(\frac{wT}{2}\right)}{wT}\right]\cdot \underset{w\to 0}{\lim_{T\to 0}}\left[\frac{\sin\left(\frac{wT}{2}\right)}{w}\right]$$ and say that the limit of the first term tends to 1, but what about the second term? Thank you!

@julien i interpret it this way, john please check if I did it right — Dominic Michaelis, Mar 24 '13 at 15:40

score 1 · Answer 1 · answered Mar 24 '13 at 15:35

1

you can use that for small values of $w$ and $T$ $$\sin\left(\frac{wT}{2}\right)\approxeq \frac{wT}{2w} $$ So the limit should be $0$

answered Mar 24 '13 at 15:35

Dominic Michaelis

19,935

score 0 · Answer 2 · answered Mar 24 '13 at 15:39

0

I'll just sequentially use limit. First in $w$ and then in $T$: $\lim_{T \to 0 , w\to 0} \frac{\sin^2(wT/2)}{w^2T}=\lim_{T \to 0 }\lim_{w\to 0}\frac{\sin(wT/2)}{wT}\frac{\sin(wT/2)}{w}=\lim_{T \to 0 }\frac{1}{2}\frac{T}{2}=0$

answered Mar 24 '13 at 15:39

Caran-d'Ache

3,564

why can you look at first the limit of $w \to 0$ and afterward the other? Some limits depend on that – Dominic Michaelis Mar 24 '13 at 15:42
@DominicMichaelis because in this case double limit can be reduced to iterated limit. It depends on the function under limit sign. – Caran-d'Ache Mar 24 '13 at 15:46
@DominicMichaelis well, for "double limit Vs iterated limit" you can refer here (http://books.google.ru/books?id=gD4RHZeYcbMC&pg=PA186&lpg=PA186&dq=double+limit+and+iterated+limit&source=bl&ots=uJXCQ3YnrI&sig=kAgHRVDJ0ZCrEXz4-a1zfJKQNWk&hl=ru&sa=X&ei=KR9PUbCbJ-Xl4QTD6oG4Bg&redir_esc=y#v=onepage&q=double%20limit%20and%20iterated%20limit&f=false): Operator Theory and Ill-Posed Problems by L. Ya: Savelev , M. M. Lavrentev, Hardcover, 696 Pages, Published 2006 by Brill Academic Publishers (Chapter 3, page 185). – Caran-d'Ache Mar 24 '13 at 15:59

C-star-W-star · Answer 3 · 2015-10-16T10:08:44.717

Problem

Double limits may heavily depend on order: $$a_{m,n}:=\frac{m}{m+n}:\quad\lim_n\lim_ma_{mn}=1\quad\lim_m\lim_na_{mn}=0\quad\lim_{n=m^2}a_{mn}=\infty$$

But it can be even more abstruse as for: $$\Delta a_{mn}=a_{mn}-a_{nm}:\quad\lim_m\Delta a_{mn}=1\quad\lim_m\lim_n\Delta a_{mn}=1\quad\lim_{n=m^2}a_{mn}\notin\mathbb{R}$$

So you certainly have to be careful about that!!

Application

Many really bad mistakes may be done in physics this way...

(Narrow Potential, Thermodynamic Limit, Bose-Einstein Condensation)

Interesting double Limit

3 Answers3