How to show that limit value exists? Here I have function 
$$f(x,y)=\frac{x^3-xy^2}{x^2+y^2}$$
I know that $\lim\limits_{(x,y) \to (0,0)}f(x,y)=0$ because if we put $y=mx$, then we have 
$$f(x,mx)=\frac{x(1-m^2)}{1+m^2}$$
$$\Rightarrow~\lim\limits_{x \to 0}f(x,mx)=0 $$
But how to show through $\epsilon-\delta$ definition that limiting value of the function exists.
Thanks in advance
 A: For $x,y\in\mathbb{R}$ is $$0\leq|x^2-y^2|\leq x^2+y^2,$$ thus if $xy \neq 0$ we get $$0\leq\frac{|x^2-y^2|}{x^2+y^2}\leq 1.$$ Then $$0<|x|\frac{|x^2-y^2|}{x^2+y^2}\leq |x|$$ and the value $0$ of the limit follows immediately.
Note added to complete $\epsilon-\delta.$ 
Since $$0<|x|\frac{|x^2-y^2|}{x^2+y^2}\leq |x|\leq \sqrt{x^2+y^2},$$ for any $\epsilon$ one can choose $\delta=\epsilon,$ and it is done.
A: I would write this equation in polar coordinates. $$x=r\cos\theta\\y=r\sin\theta$$
Then $$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{r\to 0}\frac{r^3(\cos^3\theta-\cos\theta\sin^2\theta)}{r^2}$$
The absolute value of the expression in parenthesis is always less that $2$, so it's easy to write the $\delta-\varepsilon$ solution.
A: $\left |\dfrac{x(x^2-y^2)}{x^2+y^2} \right |\le$
$|x| + \left |\dfrac{x(-2y^2)}{x^2+y^2}\right | \le $
$|x| + |2x|  = 3|x| =$
$3\sqrt{x^2} \le 3\sqrt{x^2+y^2}$.
Given $\epsilon >0.$
Choose $\delta =\epsilon/3$.
A: Showing that on y=mx curve your limit is the same for every m isn't sufficient, because u can approach the point (0,0) through, let's say, y=sin(x) curve. (sin(x) tends to 0, as x tends to 0, everything's still okay).
But not to stray away from the topic, let's see:
$$\frac{x^3-xy^2}{x^2+y^2} = \frac{(x+y)(x-y)}{x^2+y^2}x $$
Let's forget for a while about the last "x" that I've left from fraction.
Look at phrase:
$$ \frac{x^2-y^2}{x^2+y^2} = 1 - 2 \frac{y^2}{x^2+y^2} $$
Clearly, that cannot be either greater than 1 ( cause $\frac{y^2}{x^2+y^2} $ isn't negative, or be less than -1 ( cause $\frac{y^2}{x^2+y^2}$ cannot be greater than 1)
So that we get VERY important inequality:
$$ 0 \leq |\frac{x^3-xy^2}{x^2+y^2}| \leq |x| $$
Finally, let's conclude that x tends to 0 as (x,y) tends to (0,0).
By squeeze theorem ( I believe it is called so in English) you have what you wanted.
