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Consider $$a_{mn}=\frac{m^2n^2}{m^2+n^2}\left(1-\cos\left(\frac{1}{m}\right)\cos\left(\frac{1}{n}\right)\right)$$

Does $\lim_{m,n\to\infty}a_{mn}$ exist?

It can be seen that $$\lim_{m\to\infty}\left(\lim_{n\to\infty}a_{mn}\right)=\lim_{n\to\infty}\left(\lim_{m\to\infty}a_{mn}\right)=\frac{1}{2}$$

However, I still cannot determine whether the limit exists or not. Any one can help? Thanks!

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  • $\begingroup$ Can you explain to an amateur what the difference is between taking the two limits separately, and taking the limits together? $\endgroup$ – Alfred Yerger Jan 31 '15 at 23:07
  • $\begingroup$ Presumably the double limit in this case means that for all $\epsilon>0$, there exists $N>0$ such that $|a_{mn}-L|<\epsilon$ for all $m,n>N$. $\endgroup$ – Alex R. Jan 31 '15 at 23:14
  • $\begingroup$ The terms are nonnegative so by Fubini's theorem the series converges. I believe, anyway. $\endgroup$ – Math1000 Jan 31 '15 at 23:15
  • $\begingroup$ how about approximating $\cos(t) \approx 1 - \frac12 t^2$ $\endgroup$ – user66081 Jan 31 '15 at 23:20
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The answer is that the limit does exist.

Rewrite this using $u = \frac{1}{m}$ and $v = \frac{1}{n}$ as $$ a(u,v) = \frac{1}{u^2+v^2} \left( 1 - \cos(u) \cos(v) \right) $$

We are now looking for the limit as $u$ and $v$ approach zero. The limit will exist if and only if limits exist along all straight lines through the origin and these limits are all the same value.

So write $u = rv$ and define $$ L(r) \equiv \lim_{v\to 0} a(rv,v) $$

The limit exists if and only if for all $r$, $L(r)$ exists, and for $\forall r_1, r_2: L(r_1) = L(r_2)$, and $L(0) = \lim_{u\to 0} a(u,0)$ . Your work shows this third point, and in fact $L(r)$ always exists, so the key is to see whether $L(r)$ is independent of $r$.

$$ L(r)=\lim_{v \to 0} \frac{1}{v^2r^2+v^2} \left(1-\cos(rv)\cos(v)\right)\\= \lim_{v \to 0} \frac{1}{v^2} \frac{1}{1+r^2} \left( 1 - (1 - r^2v^2/2 +O(v^4) ) (1 - v^2/2 +O(v^4) )\right)\\ = \frac{1}{1+r^2}\frac{1}{2}(r^2+1)=\frac{1}{2} $$ That is independent of $r$ and matches the necessary value so the limit exists and is $\frac{1}{2}$.

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  • $\begingroup$ Why is it enough to look at limits along all lines through the origin? $\endgroup$ – user84413 Feb 1 '15 at 0:17
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We have

\begin{align}a_{mn} &= \frac{m^2n^2}{m^2 + n^2}\left(1 - \left(1 + \frac{1}{2m^2} + O\left(\frac{1}{m^4}\right)\right)\left(1 + \frac{1}{2n^2} + O\left(\frac{1}{n^4}\right)\right)\right)\\ &=\frac{m^2n^2}{m^2 + n^2}\left(1 - \left(1 - \frac{1}{2}\left(\frac{1}{m^2} + \frac{1}{2n^2}\right) + O\left(\frac{1}{m^4n^4}\right)\right)\right)\\ &= \frac{m^2n^2}{m^2 + n^2}\left(\frac{1}{2}\frac{m^2 + n^2}{m^2 n^2} + O\left(\frac{1}{m^4n^4}\right)\right)\\ &= \frac{1}{2} + O\left(\frac{1}{m^2n^2(m^2 + n^2)}\right)\\ &= \frac{1}{2} + O\left(\frac{1}{m^3n^3}\right),\quad \text{as $m,n \to \infty$} \end{align}

So there is a positive integer $k$ and a constant $C > 0$ such that $|a_{mn} - \frac{1}{2}| < \frac{C}{m^3n^3}$ for all $m, n \ge k$. Given $\varepsilon > 0$, choose $N > \max\{N, (\frac{C}{\varepsilon})^{1/6}\}$. Then $|a_{mn} - \frac{1}{2}| < \varepsilon$ for all $m, n \ge N$. Therefore, $\lim_{m,n\to \infty} a_{mn}$ exists and equals $1/2$.

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