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I would like to show that:

$ \frac{x\sin(y)-y\sin(x)}{x^2+y^2}\rightarrow_{(x,y)\to (0,0)}0$

$ \left| \frac{x\sin(y)-y\sin(x)}{x^2+y^2} \right| \leq \frac{2\vert xy \vert}{x^2+y^2} \leq 1$ which is not sharp enough, obviously.

How can I efficiently "dominate" the quantity $ \vert x\sin(y)-y\sin(x)\vert$ ?

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    Hint: When $x$ is small, \left | \sin x -x\right |.2012-03-18

3 Answers 3

0

Using polar coordinates:

$\lim_{r->0^+}\frac{r\cos(\theta)\sin(r\sin(\theta))-r\sin(\theta)\sin(r\cos(\theta))}{r^2}=\lim_{r->0^+}\frac{\cos(\theta)\sin(r\sin(\theta))-\sin(\theta)\sin(r\cos(\theta))}{r}=$

(using De L'hospital rule)

$\lim_{r->0^+}{\cos(\theta)\sin(\theta)\cos(r\sin(\theta))-\sin(\theta)\cos(\theta)\cos(r\cos(\theta))}=$ $\sin(\theta)\cos(\theta)\lim_{r->0^+}\cos(r\sin(\theta))-cos(r\cos(\theta))=0$

7

If $xy=0$ then your function is zero. You could just write for $x,y \neq 0$ $\left| \frac{x\sin(y)-y\sin(x)}{x^2+y^2} \right| = \frac{|xy|}{x^2+y^2}\left|\frac{\sin x}{x}-\frac{\sin y}{y}\right| \leq \frac{1}{2}\left|\frac{\sin x}{x}-\frac{\sin y}{y}\right|$ which tends to zero as $x,y \to 0$.

5

We have by the fundamental theorem of analysis that $x\sin y=x\int_0^y\cos tdt=x\left(y\cos y-\int_0^y-t\sin tdt\right)=xy\cos y+x\int_0^yt\sin tdt$ and similarly $y\sin x=xy\cos x+y\int_0^xt\sin tdt$ so $|x\sin y-y\sin x|\leq |xy|\cdot|\cos x-\cos y|+|x|\frac{y^2}2+|y|\frac{x^2}2$ and finally $\frac{x\sin y-y\sin x}{x^2+y^2}\leq \frac 12|\cos x-\cos y|+|x|+|y|,$ which gives the result.