# Continuity of 'vectorial' function $\frac{x^2}{y^2-1}$

given is $f(x,y) = ( \frac{y}{x^2+1}, \frac{x^2}{y^2-1} )$. I have to study the continuity of the function for$(x,y) \to (0,1)$.

First function $f_1$ is continuous, since $lim f_1 = 1/1 = 1$ so the limit exists. And the function is defined on whole $\mathbb{R}^2$.

Secon function $f_2$.. Ok, here I have some trouble: $lim f_2 = lim_{r\to 0} \frac{r^2cos^2 \phi}{r^2sin^2 \phi -1} = ??$. Well I could do $lim_{r\to 0} \frac{r^2cos^2 \phi}{r^2sin^2 \phi -1} = lim_{r\to 0} \frac{cos^2 \phi}{sin^2 \phi - \frac{1}{r^2}} = 0$ since $1/r^2 \to \infty$. But this doesn't seem alright for me...

Someone told me I should set $x = r\cdot cos \phi, y = 1 + r \cdot sin\phi$. Am I allowed to do this for y?. Isn't there a different way?

Thank you very much :)

• Consider $(h,h^3+1)$ for $h\to0$. – Michael Hoppe Jan 20 '14 at 10:01

You forgot the $y=\underline 1 + r\sin\phi$ and so you get $$\lim_{r\to0} \frac{r^2\cos^2\phi}{r^2\sin^2\phi} = \cot^2\phi$$ And since $\cot$ is not constant, $f$ is not continuous in $(0,1)$.
Consider $(h,h^3+1)$ for $h\to0$.