Evaluate $\lim_\limits{(x,y)\to (0,0))}\frac{2x^2y}{x^2+y^2}$

Question

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

So the way I approach it, it by the squeeze theorem

$$\left| \frac{2x^2y}{x^2+y^2}\right|\leq \frac{2x^2y}{x^2}=2y$$

So $$\lim_{(x,y)\to (0,0)}\left| \frac{2x^2y}{x^2+y^2}\right| \leq \lim_{(x,y)\to (0,0)}2y=0$$

Therefore $$\lim_{(x,y)\to (0,0)}\frac{2x^2y}{x^2+y^2}=0$$

But in the answer the author wrote it has a product of two functions, when one is bounded and the other goes to $0$

$$\lim_{(x,y)\to (0,0)}\frac{2x^2y}{x^2+y^2}=\lim_{(x,y)\to (0,0)}x\cdot\frac{2xy}{x^2+y^2}$$

so $x\to 0$ and $\left| \dfrac{2xy}{x^2+y^2}\right|\leq 1$ why can we say it is bounded by $1$?

Is this is the way? let assume it is bounded by $1$

$$\left| \frac{2xy}{x^2+y^2}\right| \leq 1\iff \frac{2|x||y|}{|x^2|+|y^2|}\leq 1 \iff 2|x||y|\leq |x^2|+|y^2|\iff 0\leq|x|^2-2|x||y|+|y|^2\iff 0\leq (|x|-|y|)^2$$

And this is true for all $x,y$ so it is bounded by $1$?

Answer 1

$x^{2}+y^{2}\geq 2|x|\cdot|y|$, so $\left|\dfrac{2x^{2}y}{x^{2}+y^{2}}\right|\leq \dfrac{2|x|^{2}|y|}{2|x|\cdot|y|}=|x|$ for both $x\ne 0,y\ne 0$, but this inequality is still true either $x=0$ or $y=0$.

Now $|x|\rightarrow 0$ as $(x,y)\rightarrow(0,0)$, so by Squeeze Theorem, $\dfrac{2x^{2}y}{x^{2}+y^{2}}\rightarrow 0$ as $(x,y)\rightarrow(0,0)$.

Answer 2

Note that by polar coordinates

$$\frac{2x^2y}{x^2+y^2}=2r\cos^2\theta\sin\theta\to 0$$

indeed by squeeze theorem

$$-2r \le 2r\cos^2\theta\sin\theta \le 2r $$

Answer 3

Note :

$(a+b)^2 = a^2+b^2+2ab \ge 0.$

$(a-b)^2 =a^2+b^2 -2ab \ge 0$.

$\rightarrow: $

$a^2+b^2 \ge 2|ab|$, or for $a^2 +b^2\not =0:$

$1\ge \dfrac{2|ab|}{a^+b^2}.$

Note : $ a^2+b^2 \not =0 \iff $

$a \not = 0,$ or $b \not =0.$

Answer 4

• We know that:

$-(x+y)^2 \leq 0 \leq (x - y)^2 \to -(x^2 + y^2) \leq 2xy \leq (x^2 + y^2) \to -1 \leq \dfrac{2xy}{x^2 + y^2} \leq 1$

• Thus:

$|\dfrac{2xy}{x^2 + y^2}| \leq 1$

• Hence:

Yes for any value it is bounded. And for $x,y=0$ we use the limit you stated and it is still bounded.

• Note:

You approach is ok but you should not omit abstract in the final stage.