I have tried switching to polar coordinates to no avail, I have tried using a gazillion path tests to no avail, I tried sandwiching to no avail and now I am really desperate, I even attempted an $\epsilon,\delta$ proof, with no success. I have been trying to prove/disprove the existence of the following limits:
$$1) \lim_{(x,y)\to (0,0)}\frac{\sin (xy)}{|x|+|y|} $$
$$2) \lim_{(x,y)\to (0,0)}\frac{1-\cos (xy)}{xy^2} $$
Wolfram alpha tells me the limit of the first doesn't exist, and the limit of the second is $0$
May you please give me a couple of hints and if possible a PERSONAL checklist of things you do to verify limits? I know such a list has already been mentioned on here, but feel free to add or modify.
Any help is greatly appreciated :)
EDIT: I have also tried variable substitution, although I am not very comfortable using this method because I rarely have intuition as to what substitution I should use, if possible please help me on that front too :P
Both limits are $0$. You can use the inequalities $|\sin(t)|\leq |t|$ and $1-\cos(t)\leq \frac 12 t^2$ for any $t\in\mathbb{R}$. For the first limit, $$\left|\frac{\sin(xy)}{|x|+|y|}\right|=\frac{\left|\sin(xy)\right|}{|x|+|y|}\leq\frac{|xy|}{|x|+|y|}\leq\frac{|xy|}{|x|}=|y| $$ Since $|y|\to 0$, the squeeze theorem implies the first limit is $0$ as well.
For the second one, $$\left|\frac{1-\cos(xy)}{xy^2}\right|=\frac{1-\cos(xy)}{|x|y^2}\leq\frac 12\frac{x^2y^2}{|x|y^2}=\frac{|x|}{2} $$ Again, since $|x|/2\to 0$, the squeeze theorem tells us the second limit should be $0$.