# solution verification – Some doubts in the evaluation of: limit as \$(x,y)to(0,0)\$ of \$frac{sin xy}{x+y}\$

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 $$|xy|leqfrac{1}{2}(x^2+y^2)$$, we have
$$0 leq lim_{(x,y)to(0,0)}left|frac{sin xy}{x+y}right|leqlim_{(x,y)to(0,0)}frac{|xy|}{|x+y|}leqlim_{(x,y)to(0,0)} frac{x^2+y^2}{2|x+y|}$$
Using polar coordinates it is
$$lim_{(x,y)to(0,0)}frac{|xy|}{|x+y|}=lim_{rho to 0^+} frac{rho^2}{2|rho cos theta+rho sin theta|}=lim_{rho to 0^+} frac{rho}{2|cos theta+sin theta|}$$
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
$$lim_{rho to 0^+} frac{rho}{2|cos theta+sin theta|} leq lim_{rho to 0^+} frac{rho}{2m}=0$$
And this is independent from $$theta$$. So by the squeeze theorem, the limit is $$0$$.
Here’s my doubts:

1. 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.

2. 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?

3. To say that $$|sin t| leq |t|$$ I’ve used the fact that the sin function is Lipschitz-continuous with constant $$L=1$$, that is $$|sin x – sin y| leq |x-y|$$ 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?

4. 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?