Prove a limit in two variables is $0$ I have the limit
\begin{align}
\lim_{(x,y) \rightarrow(0,0)} \frac{1-\cos(xy)}{\sqrt{x^2+y^2}(x^2+y^2)}
\end{align}
First thing I did is to use Taylor, so $$1-\cos(xy)=\frac{(xy)^2}{2} -\frac{(xy)^4}{4} + o((xy)^4)$$
Therefore the limit is $$\lim_{(x,y) \rightarrow (0,0)} \frac{1}{2} \frac{x^2y^2}{\sqrt{x^2+y^2}(x^2+y^2)}$$
Now, $\frac{y^2}{x^2+y^2} \leq 1$, therefore I have
$$\left|\frac{1}{2} \frac{x^2y^2}{\sqrt{x^2+y^2}(x^2+y^2)}\right| \leq \left|\frac{x^2}{\sqrt{x^2+y^2}}\right| \leq \frac{x^2}{|x|} = |x|$$
Since $|x| \rightarrow 0 $ as $(x,y) \rightarrow (0,0)$, the limit is $0$, correct?

It seems correct, to avoid Taylor we can use that
$$\frac{1-\cos(xy)}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{1-\cos(xy)}{(xy)^2}\frac{(xy)^2}{\sqrt{x^2+y^2}(x^2+y^2)}$$
with $\frac{1-\cos(xy)}{(xy)^2} \to \frac12$ by standard limits.
For the conclusion by squeeze theorem, we should also consider a part the case $x=0$.
