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:
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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.
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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?
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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?
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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?
Best Answer
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.