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.
 A: You know that $$\lim_{t\to 0} \frac{\cos t-1}{t^2} =-\frac{1}{2}, $$using L'Hospital, for example. This tells us that $g\colon \Bbb R \to \Bbb R$ defined by $$g(t) \doteq \begin{cases} \dfrac{\cos t-1}{t^2}, \mbox{if } t \neq 0 \\[.5em] -\dfrac{1}{2}, \mbox{if }t=0\end{cases}$$is continuous. Because of that, we can write $$\lim_{(x,y)\to (0,0)} \frac{\cos (xy)-1}{x^2y^2} = \lim_{(x,y) \to (0,0)} g(xy) = g\left(\lim_{(x,y)\to (0,0)}xy\right) = g(0) = -\frac{1}{2}.$$
A: 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. 
A: Hint. One may recall that, as $t \to 0$,
$$
\cos t \to 1,\qquad \frac{\sin t}{t} \to 1,
$$ giving, as $t \to 0$,
$$
\dfrac{-\sin^{2}t}{t^2(\cos(t)+1)}\to -\frac12.
$$
A: To expand my comments above into a somewhat contrarian answer: I would argue that the limit above, as written, does not exist.  This is because the definition of limit requires that the function be defined on a punctured neighborhood of the point in question; more specific to this case, we would say that the limit as $\langle x,y\rangle\to\langle0,0\rangle$ is $L$ iff for every $\epsilon$ there exists a $\delta$ such that for all $\langle x,y\rangle\neq\langle 0,0\rangle$ with $\left|\langle x,y\rangle-\langle0,0\rangle\right|\lt\delta$, $\left|f(x,y)-L\right|\lt\epsilon$. This fails to hold here because there are many $\langle x,y\rangle$ for which $f(x,y)$ (defined as $f(x,y) = \dfrac{\cos(xy)-1}{x^2y^2}$) is undefined — namely, all points of the form $\langle x,0\rangle$ or $\langle 0,y\rangle$.
Note that it's possible to define $f(x,y)$ at points on the axes in such a way that the function is continuous on all of $\mathbb{R}^2-\langle0,0\rangle$: just use the same trick that's given in the answers and say that $f(x,0)=-\frac12$ for $x\neq 0$, and likewise $f(0,y)=-\frac12$ for $y\neq0$. Then the limit of that function as $\langle x,y\rangle\to\langle0,0\rangle$ exists and can be found using the techniques presented in the other answers (with some careful justification).  But this isn't how the initial function $f()$ is defined.
A: For $xy\ne 0$ we have $$\frac {-1+\cos (xy)}{x^2y^2}=\frac {-2\sin^2(xy/2)}{x^2y^2} =\left(-\frac {1}{2}\right) \left(\frac {\sin (xy/2)}{(xy/2)}\right)^2 .$$ As other answerers have noted, the limit only exists if we restrict both $x$ and $y$ to non-zero values.
A: $$\lim_{t\to 0}\dfrac{\cos(t)-1}{t^2}=\lim_{t\to 0}\dfrac{\cos(t)-cos(0)}{t^2} $$
$$\lim_{t\to 0}\dfrac{\cos(t)-cos(0)}{t^2}=\lim_{t\to 0}\dfrac{-2\sin(\frac t 2)\sin(\frac t 2 )}{t^2} =\lim_{t\to 0}\dfrac{-2\sin^2(\frac t 2)}{t^2}$$
Since  $\lim_{x\to 0}\dfrac{ \sin(x)} {x}=1$ ,
$$\lim_{t\to 0}\dfrac{-2\sin^2(\frac t 2)}{t^2}=\frac {-1} 2\lim_{t\to 0}\dfrac{ \sin^2(\frac t 2)}{\frac {t^2}{4}}=\frac {-1}{2}$$
A: $$\lim_{(x,y)\to(0,0)}\frac{\cos(xy)-1}{x^2y^2}= -\lim_\limits{(x,y)\to(0,0)}\frac{2\sin^2 \frac{xy}{2}}{x^2y^2} = -\lim_\limits{(x,y)\to(0,0)} \frac{\sin^2 \frac{xy}{2}}{2\left(\frac{xy}{2}\right)^2} = \\ =-\dfrac{1}{2} \lim\limits_{t\to 0}\frac{\sin^2t}{t^2} = -\dfrac{1}{2}$$
