Im stuck in this exercise:

Study the continuity of the next function: $$f(x,y) = \begin{cases} \frac{x\sin(x^2+y^2)}{x^2+y^2}&\text{si } (x,y)\not=(0,0)\\ 0 &\text{si }(x,y) =(0,0). \end{cases}$$

I have gone this far:

$x=r\cos\theta $

$y=r\sin\theta $

so $$\frac{x\sin(x^2+y^2)}{x^2+y^2}=\frac{r\cos\theta·\sin(r^2\cos^2\theta+r^2\sin^2\theta)}{r^2\cos^2\theta+r^2\sin^2\theta}=\frac{r\cos\theta·\sin(r^2)}{r^2}=\frac{\cos\theta·\sin(r^2)}{r}$$

Then the limit when $r\to0$ $$\lim_{r\rightarrow0}\frac{\cos\theta·\sin(r^2)}{r}=\lim_{r\rightarrow0}\frac{\cos\theta·r^2}{r}=\lim_{r\rightarrow0}{\cos\theta·r}=0 \text{ for all }\theta \in [0,2\pi)$$

From here I don't know what I have to do in order to say if its continuous

  • $\begingroup$ Sorry, what's the sen function? $\endgroup$ – anonymous Jan 13 '13 at 0:45
  • 3
    $\begingroup$ It's $sin$ in spanish. $\endgroup$ – Git Gud Jan 13 '13 at 0:46
  • $\begingroup$ Fixed, it was just a typo. As @GitGud said in spainsh is written $sen$ $\endgroup$ – Alejandro Jan 13 '13 at 0:47
  • $\begingroup$ Although in general, to show that $f$ is continuous (and assuming sen is continuous), all you need to show is that $f(x,y) \to 0$ as $(x,y) \to 0$, which you've done, as best I can see. $\endgroup$ – anonymous Jan 13 '13 at 0:48
  • $\begingroup$ please note that the latex command for sin is \sin and for cos is \cos. $\endgroup$ – Ittay Weiss Jan 13 '13 at 0:49

You don't need to change to polar coordinates. Notice that at any point $(x,y)\ne (0,0)$ the function is continuous since it is a quotient of continuous functions. So, you just need to check the limit as $(x,y)$ goes to $(0,0)$. Noting that as $(x,y)$ goes to $(0,0)$ so does $x^2+y^2$, and since $\lim_{t\to 0}\sin(t)/t=1$ you may conclude that the limit of the function as $(x,y)\to (0,0)$ is equal to $0$ (note the multiplication by $x$ in the numerator). Since $f(0,0)=0$ be definition it follows that the function is continuous everywhere.

  • $\begingroup$ Would it be valid with what I have done in my question like anonymous says? or is not enough to demostrate that is continuos in the point $(0,0)$? $\endgroup$ – Alejandro Jan 13 '13 at 4:44
  • $\begingroup$ Yes, it suffices. As remarked the function is clearly continuous at any point other than (0,0) since it a quotient of continuous functions. Then you showed it is also continuous at (0,0) by means of polar coordinates. It is not wrong, but it is un-needed in this case. $\endgroup$ – Ittay Weiss Jan 13 '13 at 4:47
  • $\begingroup$ Thanks! haha looks like I asked a problem that was already solved. Ouch :) $\endgroup$ – Alejandro Jan 13 '13 at 4:59

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