For $\beta\in\Bbb R,$ define $$f(x,y)= \begin{cases}\cfrac{x^2|x|^{\beta}y}{x^4+y^2},&x\neq 0 \\ 0, &x=0 \end{cases}$$
Prove that at $(0,0)$ the function is discontinuous if $\beta=0.$
My solution goes like this:
We can prove that $f$ is discontinuous at $(0,0)$ if, $\lim_{(x,y)\to(0,0)}f(x,y)\neq f(0,0).$
We try to show that in this case, $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist.
A general strategy to show this, is to take two functions, say $y=\phi(x)$ and $y=\psi(x)$ such that $\lim_{x\to 0}\phi(x)=\lim_{x\to 0}\psi(x)=0$ and show that $\lim_{x\to 0} f(x,\phi(x))\neq \lim_{x\to 0} f(x,\psi(x))$ (, also we note that, as $x\to 0$ we have, $(x,y)=(x,\phi(x))\to(0,0)$ and $(x,y)=(x,\phi(x))\to (0,0)$).
We let $y=\phi(x)=x$ and $y=\psi(x)=x^2.$ We have, $$f(x,y)=\begin{cases} \cfrac{x^2y}{x^4+y^2}, & x\neq 0 \\ 0,&x=0 \end{cases}$$ and so, if $L=\lim_{(x,y)\to (0,0)} f(x,y)$ exists then, $L=\lim_{x\to 0}f(x,\phi(x))=\lim_{x\to 0}\frac{x^3}{x^4+x^2}=\lim_{x\to 0}\frac{x}{x^2+1}=0$.
Again, $L=\lim_{x\to 0}f(x,\psi(x))=\lim_{x\to 0}\frac{x^4}{x^4+x^4}=\lim_{x\to 0}\frac{1}{2}=\frac 12.$
This is what we've wanted to show and $L=\frac 12$ gives us a contradiction as $\frac 12\neq 0.$
So, $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist. This proves that at $(0,0)$ the function is discontinuous if $\beta=0.$
On second thought, I tried to prove the same thing using the "approach of polar coordinates' as follows:
Just like the previous approach, we try to show that $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist but this time using polar coordinates.
We have,
$\lim_{(x,y)\to(0,0)}\cfrac{x^2y}{x^4+y^2}=\lim_{r\to 0} \cfrac {r^3\cos^2\theta\sin\theta}{r^4\cos^4\theta+r^2\sin^2\theta}=\lim_{r\to 0} \cfrac {r\cos^2\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}=0.$
So, surprisingly we end up seeing that $\lim_{(x,y)\to(0,0)}f(x,y)=f(0,0)$ due to which we may conclude that $f$ is continuous at $(0,0).$
Both approaches should have given me the same conclusion, but this is not happening in my case. However, I don't understand where did I go wrong in this second approach of mine? Is my first approach, correct? Any help regarding this issue will be greatly appreciated.
Best Answer
From the purely logical standpoint:
In your first approach you found two curves that gave different values of the limit ($2$). This correctly proves that the limit ($1$) does not exist.
In your second approach you essentially proved that when the curve $(\varphi(t), \psi(t))$ is any straight line, the value of the limit ($2$) equals $0$. This does not prove that the limit ($1$) equals $0$, because you haven't considered every possible curve.
The reason why your approach with polar coordinates is equivalent to computing the limit over a straight line (rather than an arbitrary curve) is as follows: the limit
$$\lim_{r \to 0} g(r \cos \theta, r \sin \theta)$$
is a special case of the limit ($2$), where $\varphi(r) = r \cos \theta$ and $\psi(r) = r \sin \theta$. The curve that it gives, i.e. $\{ (\varphi(r), \psi(r)) : r \in \mathbb{R} \}$ is a straight line going in the direction of $(\cos \theta, \sin \theta)$.