Evaluate $\lim_{(x,y)\to(0,0)}\frac{\cos(xy)-1}{x^2y^2}$ Evaluate $\lim_\limits{(x,y)\to(0,0)}\dfrac{\cos(xy)-1}{x^2y^2}$
The limit does exist.
The only thing I can think of is let $t=xy$. Then our limit becomes
$$\lim_{t\to 0}\dfrac{\cos(t)-1}{t^2}=\lim_{t\to 0}\dfrac{\cos(t)-1}{t^2}\dfrac{(\cos(t)+1)}{(\cos(t)+1)}=\lim_{t\to 0}\dfrac{-\sin^{2}t}{t^2(\cos(t)+1)}$$
This is how far I got and then I don't know. Not even sure if this is the best method to find the limit.
Any hints would help.
Thank you.

Hint. One may recall that, as $t \to 0$,
$$
\cos t \to 1,\qquad \frac{\sin t}{t} \to 1,
$$ giving, as $t \to 0$,
$$
\dfrac{-\sin^{2}t}{t^2(\cos(t)+1)}\to -\frac12.
$$
