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?
 A: $$\frac{1 - \cos(\pi x y ) + \sin (\pi(x^2 + y^2))}{x^2 + y^2} = \frac{1 - \cos(\pi x y )  }{x^2 + y^2}+\pi\frac{ \sin (\pi(x^2 + y^2))}{\pi(x^2 + y^2)}$$
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 $\pi$
A: On the one hand, $$\lim_{r\to 0}{\frac{\sin(\pi r^2)}{r^2}}=\pi\cdot\lim_{r\to 0}{\frac{\sin(\pi r^2)}{\pi r^2}}\overset{t=\pi r^2}{=}\pi\cdot\lim_{t\to 0}{\frac{\sin t}{t}}=\pi\cdot 1=\pi.$$
On the other hand,
\begin{align*}
\lim_{r\to 0}{\frac{1-\cos(\pi r^2\cos\theta\sin\theta)}{r^2}}&\overset{t=\pi r^2\cos\theta\sin\theta}{=}\pi\cos\theta\sin\theta\cdot\lim_{t\to 0}{\frac{1-\cos t}{t}}=0,
\end{align*}
because $$\lim_{t\to 0}{\frac{1-\cos t}{t}}=\lim_{t\to 0}{\frac{2\sin^2(t/2)}{t}}\overset{s=t/2}{=}\lim_{s\to 0}{\frac{\sin^2s}{s}}=\lim_{s\to 0}{\frac{\sin^2 s}{s^2}\cdot s}=1^2\cdot 0=0.$$
Therefore, the limit is $\pi$.
A: hint
$$1-\cos(2A)=2\sin^2(A)$$
$$=2\sin^2(|A|)\le 2A^2$$
$$\lim_{r\to0}\frac{\sin(\pi.r^2)}{r^2}=\pi$$
