How to prove this two variable limit doesn't exist?

Question

$\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2-y}$

I set $x = r\cos\theta$, $y=r\sin\theta$, when $(x,y) \to (0,0)$, $r\to0$ and I got $\lim_{r \to 0} \frac{\cos^2\theta}{\cos^2\theta-\sin\theta/r}$ which is undefined.

Is it enough to prove this ?

Answer 1

Yes, your approach is enough. Perhaps that it would be better to add that $\theta\notin\pi\mathbb Z$; otherwise, the limit exists.

Answer 2

Let $f(x,y)=\frac{x^2}{x^2-y}$. Observe that $f(x,0)=1$ and $f(0,y)=0$. Therefore the limit does not exist.

Answer 3

Note that

• for $x=0 \implies \frac{x^2}{x^2-y}=0$
• for $x=t \quad y=t^2-t^3 \quad t\to 0^+\implies \frac{x^2}{x^2-y}=\frac{t^2}{t^2-t^2+t^3}=\frac1t\to\infty$