What's wrong with the method I used to prove this limit doesn't exist?

Question

The limit is

$\lim_{x\to0,y\to0}\frac{x^2y}{x^2+y^2}$

Here is the way to prove the limit is $0$:

For arbitrary $\epsilon$, there exists a $\delta $ ($ 0 < \sqrt{x^2+y^2} < \delta$) that can satisfy the equation $| {\frac {x^2y} {x^2+y^2}}|<\epsilon$ as $x^2 \le x^2+y^2$ hence $|\frac{x^2y}{x^2+y^y}| \le |y| = y^2\le \sqrt {x^2+y^2}$ if we choose $\delta = \epsilon$.

Nevertheless, I try to choose (x, y) approaching (0,0) along a particular curve such as $x^2=-y^2+y$ and $x^2=-y^2+y^3$, which give me the limits are $0$ and $-1$ respectively that can prove the limit doesn't exist.

The question I wanna ask is what's wrong the curve I chosen and what is the rule to or not to choose a particular curve.

Any help will be appreciated, thank you!

Notice that $-y^2+y^3 < 0$ for all $y<1$, except for $y=0$. This means that you cannot approach $(0,0)$ along the curve $x^2=-y^2+y^3$, since on this curve we must have $-y^2+y^3 = x^2\ge 0$. — Joey Zou, Sep 25 '17 at 02:01
@JoeyZou I think you should turn your comment into an answer, because it directly explains the mistake in OP's reasoning. — Chessanator, Sep 25 '17 at 11:36

score -1 · Answer 1 · answered Sep 25 '17 at 09:42

-1

You can do this also:

Put $$x=r\cos{t}$$ $$y=r\sin{t}$$

Then:

$$\lim_{(x,y) \to (0,0)}\frac{x^2y}{x^2+y^2}=\lim_{r \to 0} \frac{r^3 \cos^2{t}\sin{t}}{r^2}=\lim_{r \to 0}r \cos^2{t}\sin{t} = 0$$, $\forall t \in [0,2\pi)$

Thus the limit is zero.

In general it is a very useful technique to use polar coordinates for computing multivariable limits or to examine the existence of a limit.

For instanse if the result as $r \to 0$ depends on values of $t$ then the limit does not exist.

As an exercise for you try to prove that $\lim _{(x,y) \to (0,0)} \frac{x-y}{x+y}$ does not exist,using polar coordinates.

answered Sep 25 '17 at 09:42

Nikos Poursalidis

17

"The question I wanna ask is what's wrong the curve I chosen and what is the rule to or not to choose a particular curve." – AakashM Sep 25 '17 at 12:12
@Nikos Poursalidis Nikos,thank you, by the way, I want to ask another question.Actually, I learned this method on this website before, but it confused me that one used this method yet the answer in incorrect. So I guessed that it can be used while the limit exists. Thus, under what condition we can use this method. Here is the link (The first answer and second part) https://math.stackexchange.com/questions/706719/solve-lim-x-y-to-0-0-frac-ln1xy-sqrtx2y2-and-lim-x-y?rq=1 – M. Chen Sep 26 '17 at 10:48

What's wrong with the method I used to prove this limit doesn't exist?

1 Answers1