# 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 $$t\in\mathbb{R}$$ and it is $$|xy|\leq\frac{1}{2}(x^2+y^2)$$, we have $$0 \leq \lim_{(x,y)\to(0,0)}\left|\frac{\sin xy}{x+y}\right|\leq\lim_{(x,y)\to(0,0)}\frac{|xy|}{|x+y|}\leq\lim_{(x,y)\to(0,0)} \frac{x^2+y^2}{2|x+y|}$$ 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 $$[0,2\pi]$$ 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 $$|\cos \theta + \sin \theta|=0\Leftrightarrow \cos \theta =-\sin \theta$$ 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 $$\theta\in[0,2\pi)$$; here I have heavily used that $$\theta\in[0,2\pi]$$, is this the same? If yes, why can I extend the interval to $$[0,2\pi]$$ 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 $$x,y\in\mathbb{R}$$ 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?

• The minimum of $|\cos\theta+\sin\theta|$ on $[0,2\pi]$ is zero, achieved for example at $3\pi/4$. Your function is not defined along the line $x+y=0$, and $|\cos\theta+\sin\theta|$ can take values as close to zero as you want on the domain. Nov 27, 2020 at 1:57
• 2. You haven’t heavily used the fact that $\theta\in[0,2\pi]$, the interval $[0,2\pi)$ differs from a compact set by just one point. And by the way, considering “the same angle” twice can’t make you lose any information. 3. Of course you can choose $y=0$, that’s what “for all $x,y\in\mathbb{R}$” means. Nov 27, 2020 at 2:08

If you approach the origin along the line $$xy=x+y,\, (x,y)\neq (0,0)$$ (which is contained in the domain), your function approaches $$1$$ $$\left(\lim\limits_{t\to 0}\dfrac{\sin t}t=1\right)$$, and along $$x=0$$ the limit is zero. Therefore your limit does not exist.