Proving Continuous Extension of $f(x,y)$ Function

[laura1231]

Can I extend the function $f(x,y)=(x^2+y^2)\arctan\dfrac{1}{|xy|}$ to a continuous function?
If I consider the restriction of $f$ along the line $x=k$ i find $\lim_{(x,y)\rightarrow(k,0)}(x^2+y^2)\arctan\dfrac{1}{|xy|}=k^2\dfrac{\pi}{2}$
how can i prove that?

 

[I like Serena]

Can I extend the function $f(x,y)=(x^2+y^2)\arctan\dfrac{1}{|xy|}$ to a continuous function?
If I consider the restriction of $f$ along the line $x=k$ i find $\lim_{(x,y)\rightarrow(k,0)}(x^2+y^2)\arctan\dfrac{1}{|xy|}=k^2\dfrac{\pi}{2}$
how can i prove that?

Hey laura123!

To extend $f$ to a continuous function on $\mathbb R^2$ we need that for every $x_0$ the limit $\displaystyle\lim_{(x,y)\to(x_0,0)} f(x,y)$ exists. And we need the same thing for $y$ values, although that follows by symmetry.

We can try to find such a limit by switching to local polar coordinates.
That is:
$$\lim_{(x,y)\to(x_0,0)} f(x,y)
= \lim_{r\to 0} f(x_0+r\cos\phi, r\sin\phi)
$$
Can we find that limit? (Wondering)

 

[laura1231]

$$\displaystyle\lim_{(x,y)\rightarrow(x_0,0)}(x^2+y^2)\arctan\dfrac{1}{|xy|}=$$
$$=\displaystyle\lim_{r\rightarrow0}[(x_0+r\cos\phi)^2+(r\sin\phi)^2]\arctan\dfrac{1}{|(x_0+r\cos\phi)r\sin\phi|}=$$
$$=\displaystyle\lim_{r\rightarrow0}(x_0^2+r^2\cos^2\phi+2rx_0\cos\phi+r^2\sin^2\phi)\arctan\dfrac{1}{|x_0r\sin\phi+r^2\cos\phi \sin\phi|}=$$
$$=\displaystyle\lim_{r\rightarrow0}(x_0^2+2rx_0\cos\phi+r^2)\arctan\dfrac{1}{|r(x_0\sin\phi+r\cos\phi \sin\phi)|}=x_0^2\frac{\pi}{2},\ \ \forall\phi\in]0;2\pi],\phi\neq\pi$$

 

[I like Serena]

Yep. (Nod)
So we can indeed extend f to a continuous function with $f(x,0)=x^2\frac\pi 2$, $f(0,y)=y^2\frac\pi 2$, and $f(0,0)=0$.