I must evaluate
$$lim_{(x,y)to(0,0)}frac{sin xy}{x+y}$$
My reasoning is the following, can someone tell me if this is correct?
Since $sin t leq t$ for all $tinmathbb{R}$ and it is $xyleqfrac{1}{2}(x^2+y^2)$, we have
$$0 leq lim_{(x,y)to(0,0)}leftfrac{sin xy}{x+y}rightleqlim_{(x,y)to(0,0)}frac{xy}{x+y}leqlim_{(x,y)to(0,0)} frac{x^2+y^2}{2x+y}$$
Using polar coordinates it is
$$lim_{(x,y)to(0,0)}frac{xy}{x+y}=lim_{rho to 0^+} frac{rho^2}{2rho cos theta+rho sin theta}=lim_{rho to 0^+} frac{rho}{2cos theta+sin theta}$$
Let $g(theta):=cos theta+sin theta$, $g$ is a continuous function and the interval $(0,2pi)$ is compact, so for Weierstrass it has a minimum $m>0$. So
$$lim_{rho to 0^+} frac{rho}{2cos theta+sin theta} leq lim_{rho to 0^+} frac{rho}{2m}=0$$
And this is independent from $theta$. So by the squeeze theorem, the limit is $0$.
Here’s my doubts:

I said that $m>0$ and not just $m geq0$ because $cos theta + sin theta=0Leftrightarrow cos theta =sin theta$ and this is not allowed because $frac{sin xy}{x+y}$ is not defined for those values, but I’m not sure if this is correct because I am in a limit context.

I know that polar coordinates, to be bijective, must be such that $thetain(0,2pi)$; here I have heavily used that $thetain(0,2pi)$, is this the same? If yes, why can I extend the interval to $(0,2pi)$ and not lose informations because now polar coordinates aren’t bijective anymore?

To say that $sin t leq t$ I’ve used the fact that the sin function is Lipschitzcontinuous with constant $L=1$, that is $sin x – sin y leq xy$ for all $x,yinmathbb{R}$ using this inequality with $0$ in place of $y$ since it is valid for all $y$. Is this correct or I’m not allowed to fix one of the two variables?

Why using polar coordinates independent from the angle $theta$ shows that the limit surely exists? If I’m not wrong, $theta$, being an angle, represents only the lines passing through the origin and not all the possible paths; so I’m a bit confused because with polar coordinates I’ve eliminated only the possibilities of line paths. Can someone clarify this please?