Proving continuity at $0$ for $f(x,y)=\dfrac{x^2y}{x^2+y^2}$

Question

Here is my attempt at proving the problem below.

$$\bigg|\dfrac{x^2y}{x^2+y^2}\bigg| \leq \bigg|\dfrac{x^2y}{2xy}\bigg| \text{ by the hint}$$

$$=\bigg|\dfrac{x}{2}\bigg| \leq |x| $$

Since we're using the delta epsillon definition to prove continuity, it should then follow that $|x|\leq|(x,y)|$. Have I missed anything or done something incorrectly here?

Thanks!

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Answer 1

Yes, it is correct. $f(x,y)=0$ if $(x,y) \neq (0,0)$, so in order to prove the continuity at $(0,0)$ you have to prove that

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

By the $AM-GM$ inequality we have that

$$\frac{x_1+...+x_n}{n} \geq (x_1 \cdot ... \cdot x_n)^{1/n}$$

So taking $x_1 =x^2 , x_2=y^2$:

$$\frac{x^2+y^2}{2} \geq (x^2 y^2)^{1/2} \implies x^2 + y^2 \geq 2 |xy| $$

And as you have proved

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

The absolute value of the component of a vector is always less or equal than the norm of the vector, so

$$\bigg|\dfrac{x^2y}{x^2+y^2}\bigg| \leq |x| \leq \|(x,y)\|$$

Now let $\epsilon \gt0$. Taking $\delta = \epsilon$:

$$\text{if $(x,y)$ is such that } 0 \lt \|(x,y)-(0,0) \| \lt \delta \implies \bigg|\dfrac{x^2y}{x^2+y^2} - 0 \ \bigg| \lt \epsilon$$

Answer 2

Since $|x|\to 0$ when $|(x,y)|\to 0$, we have the result. In this case, you don't need the hint. Just use $$ \left|\frac{x^2}{x^2+y^2}\right| \leq 1 $$ so $$ \left|\frac{x^2y}{x^2+y^2}\right| \leq 1\cdot |y| \to 0 \qquad \text{if $|(x,y)|\to 0$} $$