Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was working on this problem and I need to show that

Let $f:\mathbb R^2 \rightarrow \mathbb R$ be defined as follows:

$f(x,y)=\frac{x^2y}{x^4+y^2}$ for $(x,y)\neq(0.0)$ and $f(0,0)=0$.

How do I show rigorously that $\frac{\partial f}{\partial y}$ is not bounded.

I kind of have a fair idea as to why it is not bounded. You would simply compute the partial and see how the degrees of the two polynomials behave in the denominator and numerator. But I am not sure if this is acceptable.

Also, is there a general criteria in showing that a function of two variables is not bounded?. Can just show that the limit goes to infinity as $(x,y)$ goes to infinity? This is something that has escaped me when I learned multivariable calculus. Can anyone help me?.

Thanks for your help.

share|cite|improve this question
up vote 3 down vote accepted

Here is the definition of a function from $\mathbb{R}^2$ to $\mathbb{R}$ being unbounded:

$f:\mathbb{R}^2\to\mathbb{R}$ is unbounded if for every $C \geq 0$ there exists $(x,y)\in\mathbb{R}^2$ such that $|f(x,y)| \geq C$.

This is all you need to work this problem out. There is no need to do anything with degrees of polynomials.

Like you suggested, let's compute the partial derivative with respect to $y$:

$$ \frac{\partial f}{\partial y} = \frac{x^2}{x^4+y^2} - \frac{2x^2y^2}{(x^4+y^2)^2}$$

Can we make $\left|\frac{\partial f}{\partial y}\right|$ larger than any $C \geq 0$? Hint: how does $\left|\frac{\partial f}{\partial y}\right|$ behave along the line $y=0$?

share|cite|improve this answer
So $\vert \frac{\partial f}{\partial y}\vert=1/x^2$ along the line $y=0$. So this goes to zero as $x$ goes to infinity?, is that correct? – Jack Dawkins Jan 14 '13 at 6:38
@user54755 Well, it does, but you wanted to prove that it's unbounded...try letting $x \to 0$ instead. – Gyu Eun Lee Jan 14 '13 at 6:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.