Does the limit of this double sequence exist? 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!

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 $1/2$.
