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Lets study the limit

$$ \lim_{(x,y)\to(0,0)} \frac{x^3 y}{x^6 + y^2} $$

If we look at the limit along any straight line eg $y = mx$ we find that the limit tends to $0$.

Studying the limit closer we test for every curve $y = x^k$, where $k$ is some integer number. This gives that for every $k\neq 3$, the limit tends to $0$. If $k=3$, the limit tends to $1/2$.

Now this shows that no matter what straigt line, or curve (except $y=x^3$) the limit tends to zero. I also tried a few other polynomials and they all tend to zero.

My assumption is the following:

The limit $$ L = \lim_{(x,y)\to(0,0)} \frac{x^3 y}{x^6 + y^2} $$ Is equal to zero if $y$ is any polynomial except $y = m x^3\,,\ x \in \mathbb{R}$.

Is the claim true? If so can anyone help proving it?

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Related: – Jack Feb 6 '13 at 16:53
Got something from an answer below? – Did Feb 9 '13 at 10:14
up vote 1 down vote accepted

It's not true. Take for example $y=x^3+x^4$. Then

$$f(x,y) = \frac{1+x}{2+2x+x^2}$$

and the limit along that curve will be $1/2$.

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Indeed! But what curves then will tend to something different than $0$? – N3buchadnezzar Feb 6 '13 at 16:37
For curves $y=f(x)$, where $f(x)=ax^3+o(x^3)$, and $a\ne 0$, the limit is not equal to $0$. I do not know a full answer. – André Nicolas Feb 6 '13 at 16:57

The set of limit points is exactly $[-\frac12,\frac12]$. In particular, the limit at $(0,0)$ does not exist.

Every value in $[-\frac12,\frac12]$ is a limit point since, for every fixed $a$, $f(x,ax^3)=\frac{a}{a^2+1}$ for every nonzero $x$.

On the other hand, for every $(x,y)$, $2|x^3y|\leqslant x^6+y^2$ hence $|f(x,y)|\leqslant\frac12$ for every $(x,y)\ne(0,0)$.

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