The limit of $f(x,y)= \dfrac {x^2 y}{x^2 + y^2}$ as $ (x,y) \to (0,0)$

Question

In order to prove that the limit as $(x,y)$ approaches to $(0,0)$ of $f(x,y)= \dfrac {x^2 y}{x^2 + y^2}$ is equal to $0$ is wanted to proof:

for ever $\beta\gt0$ exists some $\delta\gt0 $such that for ever $(x,y)$ : $ 0<\sqrt{(x^2 + y^2)}\lt\delta\implies \lvert \frac {(x²y)}{(x²+y²)}\rvert \lt \beta$

i found that $\lvert x² + y²\rvert\lt \delta^2$ and that $\lvert x²y\rvert\lt \delta^3$, but i know conclude why should $\lvert \frac {(x²y)}{(x²+y²)}\rvert \lt \beta$

How can it be done? or how can we formally proof the limit, in another way? Thank you.

Answer 1

If $ \;0< \sqrt {x^2+y^2}<b\; $ then $\;|y|<b \;$ so $\;|x^2y/(x^2+y^2)|=|x^2/(x^2+y^2)|\cdot |y|\leq |y|<b. $

Answer 2

Note that for $x \neq 0$ we have

$$ |f(x,y)| = \left| \frac{x^2y}{x^2 + y^2} \right| = \left| \frac{y}{1 + \left( \frac{y}{x} \right) ^2} \right| \leq |y| \leq \sqrt{x^2 + y^2}. $$

This inequality also holds if $x = 0$ and so if $\sqrt{x^2 + y^2} < \beta$ then $|f(x,y)| \leq \beta$ showing that $\lim_{(x,y) \to (0,0)} f(x,y) = 0$.

Answer 3

Use polar cordinates : $x=r\cos { \alpha } \\ y=r\sin { \alpha } \\ 0\le r\le +\infty $ $$\\ \\ \lim _{ r\rightarrow 0 }{ f\left( r\cos { \alpha } ,r\sin { \alpha } \right) } =\lim _{ r\rightarrow 0 }{ \frac { { r }^{ 3 }\cos ^{ 2 }{ \alpha \sin { \alpha } } }{ { r }^{ 2 }\cos ^{ 2 }{ \alpha } +{ r }^{ 2 }\sin ^{ 2 }{ \alpha } } = } \lim _{ r\rightarrow 0 }{ \frac { { r }^{ 3 }\cos ^{ 2 }{ \alpha \sin { \alpha } } }{ { r }^{ 2 } } = } 0\\ $$

since in $\alpha =\frac { k\pi }{ 2 } ,k\epsilon \mathbb Z_0$ $\\ \\ \lim _{ r\rightarrow 0 }{ f\left( r\cos { \alpha } ,r\sin { \alpha } \right) } \equiv 0\\ $ in these values of $\alpha $ $$\lim _{ r\rightarrow 0 }{ f\left( r\cos { \alpha } ,r\sin { \alpha } \right) } =0=f\left( 0,0 \right) $$