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.

You are basically done,
I would break the expression inside the limit into,
$$\frac{-1}{1+\cos t} \frac{\sin t}{t} \frac{\sin t}{t}$$
Then use the product rule for limits and the famous limit $\frac{\sin t}{t} \to 1$ as $t \to 0$.

Edit
As per the answer of @Steven Stadnicki, your limit is only equal to the single variable  limit if $xy \neq 0$ as we approach the origin. Otherwise, the expression is not even defined even if we are very close to the origin. It's safer to say that $f(x,y)=\frac{\cos (xy)-1}{(xy)^2}$ tends to negative a half as we approach the origin going through points which are a part of the domain of $f$ in which the function is defined. 
