# Finding the limits of a multivariable function $\frac{x^2y^2}{x^4+y^2}$

Question: Given the following function, determine whether the following function is continuous at $$(0,0)$$
$$f(x,y)=\begin{cases}\frac{x^2y^2}{x^4+y^2}, &x^2+y^2 \neq 0,\\ 0 ,&x^2+y^2=0. \end{cases}$$

I know of three methods for approaching such problems:

1. rewrite the function in terms of polar coordinates, then try to find the limit,

2. bound the function from above and below, then apply the squeeze theorem, and

3. take limits along the line $$y=x$$.

Using the first technique of rewriting the function in terms of polar coordinates, it appears that the function is continuous at $$(0,0)$$. I would like to be able to answer the question using the squeeze theorem. I can bound the numerator, but I don't know how to bound the denominator so that the limit goes to zero at the origin.

• Do you require further help with this? Oct 15 '14 at 23:58
• I managed to get it. Thanks for ur help Oct 16 '14 at 2:28
• See here for a general case. Dec 3 '19 at 16:18

Hint: $\forall (x,y)\in \mathbb R^2\left((x,y)\neq (0,0)\implies \left|\dfrac{x^2y^2}{x^4+y^2}\right|\leq x^2\right)$.

This hint gives you yet another method which is $\varepsilon$-$\delta$.

As for approaching the limit through some paths, that only allows you to prove that limits do not exist.

• Care to comment, down voter? Dec 3 '14 at 11:33

Use the parametrization: $u = \frac{y}{x}$, $v = x$

Note that if $x^2+y^2=0$ than $(x,y) = (0,0)$, and the parametrization is defined only for $v = x \neq 0$

The function becomes: $$f(u,v)=\begin{cases}\frac{v^2u^2}{u^2+v^2}, & v \neq 0\\0 ,&v=0 \end{cases}$$

this one can be solved with polar coordinates... :)