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I have a question. I have to check whether the limit exist. But I want to check it with a) two paths or b) polar coordinates, but it doesn't work. It is about $\lim_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^3}$. Can anyone help with to show that the limit does not exist?

Thank you

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    I edit your question. Is that what you mean? – 2017-01-27
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    hint use polar coordniates and observe what happens when the polar angle is $\pi/2$ or $3\pi/2$ – 2017-01-27
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    Hint: Try to find sequences with $f(x_n, y_n)\equiv 0$ and $f(x_n, y_n)\equiv 1$ – 2017-01-27
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    Your expression is undefined on the curve $y=-|x|^{2/3}$. – 2017-01-27

1 Answers 1

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Let $f(x,y) := \frac{xy^2}{x^2+y^3}$. Then we have

$$f(x,y) = 1 \Leftrightarrow xy^2 = x^2+y^3 \Leftarrow x=\underbrace{\frac{y^2+\sqrt{y^4-4y^3}}2}_{=:h(y)}$$

Let $y_n:=1/n, x_n:=h(y_n)$. Then $f(x_n, y_n)\equiv 1$ and $$\lim_{n\to\infty}x_n = \lim_{n\to\infty}h(y_n) = h(\lim_{n\to\infty}x_n)= h(0) = 0\text{,}$$since $h:(-\infty,4)\to \mathbb R$ is continuous. Thus, $(x_n,y_n)\to_{n\to\infty}(0,0)$.

Also $f(0, 1/n) \equiv 0$.

So we have $\lim_{n\to\infty} f(x_n, y_n)\neq\lim_{n\to\infty} f(0, 1/n)$, and thus $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist.

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    As a general technique, you can always try $f(x,y)\equiv c$ or $f(x,y)=1/y$ or $f(x,y)=1/x$ and see where that takes you. – 2017-01-27
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    It seems that $y_n=−1/n$ instead of $1/n$, since x=h(y) is not real number if $y \in (0,1)$. – 2017-10-26