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Let $f(x,y)= \frac{x^2y}{\sqrt{x^6+2y^2}}$ when $ (x,y)\neq (0,0)$, $f(x,y) = 0$ otherwise. The hint says that we can prove discontinuity at $0$ by considering the sequence $(\frac{1}{n},\frac{1}{n^3})$. Even if we take this sequence, the function on this sequence converges to $0$ as $n$ becomes large. So we do not get any contradiction. $f(\frac{1}{n},\frac{1}{n^3})$ is actually $\frac {1}{\sqrt{3}n^2}$ which converges to $0$. How can we prove the discontinuity at the origin?

User2018
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    @Luchoming You may want to think about the value of that fraction when $n$ is $1000$, say. – Mark S. Jan 31 '23 at 14:17
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    The hint is wrong; the function can be extended continuously to the origin, with $f(0,0)=0$, because you can write $f(x, y) =x^2\frac{y}{\sqrt{x^6+2y^2}}$, and the fraction is bounded by $1$. – peek-a-boo Jan 31 '23 at 14:21
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    Seems that the limit is $0$ – Sine of the Time Jan 31 '23 at 15:36
  • Thanks to all, this was pestering me for quite a while. I tried all known techniques for proving discontinuity but failed. I should have observed that the function is continuous. – User2018 Jan 31 '23 at 15:41
  • When $|x| < 1,$ then $x^2 > x^6,$ so $x^6 + 2y^2 \geq x^2 + y^2,$ and therefore $f(x,y) \leq x^2 |y| / |(x,y)| \leq x^2.$ – William M. Jan 31 '23 at 20:39

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The limit at the origin is $0$, in fact:

$$\left| \frac{x^2y}{\sqrt{x^6+2y^2}}-0\right|\leq \frac{x^2|y|}{\sqrt{x^6+y^2}}\leq \frac{x^2|y|}{\sqrt{y^2}}=x^2$$ Since $g(x)=x^2$ is positive and approaches $0$ as $x\to 0$, you can conclude that the limit is $0$.

Sine of the Time
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