# Limits of multivariable functions hints

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

• For the first one, did you not notice that, at (0, 0), the numerator is 1 while the denominator is 0? For the second notice that it is $\frac{1- cos(xy)}{xy}\frac{1}{y}$ Commented Feb 12, 2020 at 12:51
• Typo, it is supposed to be $\sin$ not $\cos$. Will edit @user247327
– Kam
Commented Feb 12, 2020 at 12:52
• @user247327 I'm assuming then that $\frac{1-\cos(xy)}{xy} \to 0$ when we say $t=xy$, but isn't that just proving the limit exists with one path? Which we can't do?
– Kam
Commented Feb 12, 2020 at 12:56

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 $$0$$.
• @Kam Are you familiar with Taylor series expansions? Consider the expansion of $1-\cos(t)$ and you'll see where it comes from. It can also be proved using the derivative of $f(t)=t^2/2+\cos(t)-1$. Commented Feb 12, 2020 at 19:39