<s>[INST] Calculating $\lim_{(x,y)\rightarrow (0,0)}\frac{1 - \cos(\pi x y ) + \sin (\pi(x^2 + y^2))}{x^2 + y^2}$ if it exists I'm trying to calculate the above limit. I ran a few paths and found out that the limit is $\pi$ (which I also confirmed through WolframAlpha), but to prove it I use polar coordinates to get an expression of the form $F(r) \cdot G(\theta)$ where $F(r)\rightarrow$ when $r\rightarrow 0$ and $G$ is blocked. So far I have:
$$\lim_{r\rightarrow 0} \left|f(r\cos \theta,r \sin \theta)\right| = \lim \left|\frac{1-\cos(\pi r^2 \cos\theta \sin \theta ) + \sin (\pi r^2 (\sin^2\theta  + \cos ^2 \theta ))}{r^2\cos ^2 \theta + r^2 \sin^2 \theta}\right|
\\ \underset{\sin ^2 \theta + \cos ^2 \theta = 1}{=} \lim \left|\frac{1 - \cos(\pi r^2 \cos\theta \sin\theta ) + \sin (\pi r^2 )}{r^2}\right|$$
However, I don't know how to separate $r$ from the rest of the expression at this stage. But it occurred to me that this only works if the limit of the function is zero. So, how do I confirm this is the correct limit in a case such as this?
 [/INST] $$\frac{1 - \cos(\pi x y ) + \sin (\pi(x^2 + y^2))}{x^2 + y^2} = \frac{1 - \cos(\pi x y )  }{x^2 + y^2}+\pi\frac{ \sin (\pi(x^2 + y^2))}{\pi(x^2 + y^2)}$$
For first
$$\left|\frac{1 - \cos(\pi x y )  }{x^2 + y^2}\right|\leqslant\left|\frac{2\sin^2(\pi \frac{x^2 + y^2}{2} )  }{x^2 + y^2}\right| \to 0$$
While second is $\pi$
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