How to show that limit value exists?

Question

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

Answer 1

$\left |\dfrac{x(x^2-y^2)}{x^2+y^2} \right |\le$

$|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 $\delta =\epsilon/3$.

Answer 2

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.

Answer 3

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.

Answer 4

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.