For what values of $m$ is $\frac{y^{m}(1-\cos x)}{x^{4}+2y^{2}}$ continuous at $(0,0)$?

Question

Let $$f\left(\begin{array}{c} x\\ y \end{array}\right)=\begin{cases} \frac{y^{m}\left(1-\cos x\right)}{x^{4}+2y^{2}} & \left(\begin{array}{c} x\\ y \end{array}\right)\neq\left(\begin{array}{c} 0\\ 0 \end{array}\right)\\ 0 & \left(\begin{array}{c} x\\ y \end{array}\right)=\left(\begin{array}{c} 0\\ 0 \end{array}\right) \end{cases}$$ Where $m$ is a non-negative integer. I was asked to determine for what values of $m$ is this function continuous at the point $\left(\begin{array}{c} 0\\ 0 \end{array}\right)$. I've managed to prove it's not for $m=0$ and $m=1$ showing that for the paths defined by $\gamma\left(t\right)=\left(\begin{array}{c} t\\ t \end{array}\right)$ and $\gamma\left(t\right)=\left(\begin{array}{c} t\\ t^{2} \end{array}\right)$, the composition $f\left(\gamma\left(t\right)\right)$ is discontinuous at 0 for $m=1$ and $m=2$, respectively.

I was prepared to continue this line of reasoning for all integers, by induction, but it seems that it does not work for $m\geq 2$. On the other hand, I haven't been able to prove that $f$ is continuous for $m\geq 2$, either.

Any help would be appreciated

Answer 1

$1-\cos x \leq x^2/2 $, $\sqrt{2x^4y^2}=\sqrt{2}x^2|y|\leq\frac{x^4+2y^2}{2}$ $$| \frac{y^m(1-\cos x)}{x^4+2y^2}|\leq \frac{1}{2}\frac{|y|^mx^2}{x^4+2y^2}\leq\frac{1}{4\sqrt{2}}|y^{m-1}|$$ So for $m>1$ the function is continuous, because it implies $$ \lim_{(x, y)\to (0, 0)} \frac{y^m(1-\cos x)}{x^4+2y^2} = 0 $$

Answer 2

For $m>2$$$\lim_{(x,y)\to(0,0)}=\lim_{r\to0}\dfrac{r^{m-2}\sin^m\theta}{r^2\cos^4\theta+2\sin^2\theta}(1-\cos r\cos\theta)$$whenever $\sin\theta=0$ this limit is zero if not we have$$\lim_{r\to0}\dfrac{r^{m-2}\sin^m\theta}{r^2\cos^4\theta+2\sin^2\theta}(1-\cos r\cos\theta)=\lim_{r\to0}\dfrac{r^{m-2}\sin^m\theta}{2\sin^2\theta}(1-\cos r\cos\theta)=\lim_{r\to0}{r^{m-2}\sin^{m-2}\theta}(1-\cos r\cos\theta)=0$$then for $m\ge 3$ this function is continuous