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I want to prove that the function

$f(x,y) = \dfrac{x^2+y^2}{\sin(\sqrt{x^2+y^2})}$ if $ 0 < \sqrt{x^2 + y^2} < \pi$ and $0$ if $(x,y) = (0,0)$ is continuous at $(0,0)$ but I can't really figure it out. Can you help me?

Iuli
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Luna
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  • What is $\lim_{t\to 0} \frac{t}{\sin(t)}$? – user251257 Oct 18 '16 at 19:14
  • $\sin z$ is bounded between $\frac{2}{\pi}z$ and $z$ for $z\in\left(0,\frac{\pi}{2}\right)$, hence your function behaves like $C\sqrt{x^2+y^2}$ in a neighbourhood of the origin and $$\lim_{(x,y)\to(0,0)}f(x,y)=0.$$ – Jack D'Aurizio Oct 18 '16 at 19:30

2 Answers2

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Hint: you can use polar coordinates: $x=r\cos\theta$ and $y=r\sin\theta$ to rewrite the function $f$. Then let $r\to 0$.

Paul
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  • Thanks for your comment! Is it also possible without using polar coordinates? – Luna Oct 18 '16 at 19:16
  • @Luna: It's messier that way, but it is possible. The essence of it is to show that given any $\delta > 0$, it is possible to find an $\varepsilon > 0$ such that for all $x, y$ such that $|x|, |y| < \varepsilon$, $f(x, y) < \delta$. Even there, though, the easiest way is to fit that square in a circle and in essence use the circumscribing circle as the boundary, and using polar coordinates within that circle. – Brian Tung Oct 18 '16 at 19:30
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Let put

$x=rcos(t)$

and

$y=rsin(t)$

thus

$f(x,y)=\frac{r^2}{sin(r)}$

since

$$\lim_{r\to0}\frac{sin(r)}{r}=$$

$$\lim_{r\to 0}\frac{r}{sin(r)}=1$$,

we have

$$\lim_{(x,y)\to (0,0)}f(x,y)=$$

$$\lim_{r\to 0}r\frac{r}{sin(r)}=0$$

$=f(0,0)$.

$f$ is continuous at $(0,0)$.

hamam_Abdallah
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