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Given $$\lim_{(x,y)\to(1,\pi)}\frac{\cos(xy)}{1-x-\cos y}$$ I have reason to believe this limit actually holds and results in $-1$, as I couldn't find a counter example. Yet I find it very hard to prove this with the multidimensional version of $\epsilon - \delta$.

If I suspect the limit will be $1$, what should I be doing?

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    There's no need for using $\epsilon$ and $\delta$? Because both the limits of both the denominator and the numerator exist? – 2017-02-08
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    We know $$\lim_{(x,y)\to(1,\pi)}\frac{\cos(\pi)}{1-1-\cos\pi}=\frac{-1}{1}=-1$$ but if anyone ask that show this value is the limit we have to use $\epsilon-\delta$. – 2017-02-08

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I'll take for granted the following general property of continuous functions (it's a nice exercise):

If $f,g:\ D\ \longrightarrow\ \Bbb{R}$ are continuous at some point $d\in D$, and $g(d)\neq0$, then the function $h(x):=\frac{f(x)}{g(x)}$ is continuous at $d$.

Here we have $D=\Bbb{R}^2$ and $d=(1,\pi)$ and our functions are $$f(x,y)=\cos(xy)\qquad\text{ and }\qquad g(x,y)=1-x-\cos y,$$ where indeed $g(1,\pi)=1\neq0$. It follows that their ratio is continuous at $d$, and hence that $$\lim_{(x,y)\to(1,\pi)}\frac{\cos(xy)}{1-x-\cos y}=\frac{\cos(1\cdot\pi)}{1-1-\cos\pi}=-1.$$