(0,0) is not accumulaton point of $N_1=\{(x,y)\in\mathbb{R}^2:\frac{2x^2y}{x^2+y^2}=1\}$

Question

How do I prove that (0,0) is not accumulation point of the set $N_1=\{(x,y)\in\mathbb{R}^2:\frac{2x^2y}{x^2+y^2}=1\}$?

Answer 1

Hint:   $y \gt 0$ for all $\forall y \in N_1\,$, then (using AM-GM for the inequality step):

$$ \frac{2x^2y}{x^2+y^2}=1 \;\;\iff\;\; 2 = \frac{x^2+y^2}{x^2y} = \frac{1}{y}+\frac{y}{x^2} \ge 2 \sqrt{\frac{1}{x^2}} = \frac{2}{|x|} \;\;\implies\;\;|x| \ge 1 $$

Answer 2

For each $r>0$, let $B_{r}=\{(x,y)\mid x^{2}+y^{2}<r^{2}\}\setminus\{(0,0)\}$. We show that $B_{\frac{1}{3}}\cap N_{1}=\emptyset$. For, let $(x,y)\in B_{\frac{1}{3}}$ be arbitrary. Then there exist $0<r<\frac{1}{3}$, $\theta\in\mathbb{R}$ such that $x=r\cos\theta$ and $y=r\sin\theta$. Observe that $$ \frac{2x^{2}y}{x^{2}+y^{2}}=2r\cos^{2}\theta\sin\theta\neq1. $$ Therefore $(x,y)\notin N_{1}$. It follows that $B_{\frac{1}{3}}\cap N_{1}=\emptyset$. In particular, $(0,0)$ is not an accumulation point of $N_{1}$.

Answer 3

If $(x,y) \in N_1$, then $x \ne 0$ and

$2x^2y=x^2+y^2 \ge x^2$.

Hence $2y \ge 1$. This gives $y \ge 1/2$.

Answer 4

$x \ne 0$, $y \ne 0$; $(x,y) \in N_1$ $\rightarrow y\gt 0$.

$\star)$ $x^2 + y^2 \ge |2xy| $, since

$( x \pm y )^2 \ge 0$.

$1 = \dfrac{2x^2y}{x^2+y^2} \le \dfrac{2x^2y}{|2xy|} \le |x|$ .

Answer 5

Suppose that $(0,0)$ is an accumulation point of $N_1$ thus exists a sequece $z_n \in N_1$ such that $z_n=(x_n,y_n)$ and $z_n \rightarrow (0,0)$ and $z_n \neq (0,0), \forall n \in \mathbb{N}$

From this (using the usual metric of $\mathbb{R}^2$) it is easy to deduce that $$x_n \rightarrow 0$$ $$y_n \rightarrow 0$$

Thus $1=\frac{x_n^2y_n}{x_n^2+y_n^2} \rightarrow 0$ which is a contradiction.

Note that $$\lim_{(x,y) \rightarrow (0,0)} \frac{x^2y}{x^2+y^2}=0$$

You can prove this an an exercise if you want.