Show that $ \frac{x^2}{(x^2+y^2)^{3/4}} $ is continuous on $\mathbb{R}^2$.

Question

I tried with polar coordinates, I found the limit is infinite (which is not true the graph show it's 0). I tried to majorate with something who has for limit 0 but still impossible.

with polar coordinates : For $r > 0$ and $t \in [0,2\pi[ $, let $x=r\cos(t)$ and $y=r\sin(t)$, we have $$\frac{x^2}{(x^2+y^2)^{3/4}}=\frac{r^2\cos(t)^2}{(r^2\cos(t)^2+r^2\sin(t)^2)^{3/4}}=\frac{r^2\cos(t)^2}{(r^2)^{3/4}}=\frac{r^2}{r^{6/4}}\cos(t)^2=\sqrt{r}\cos(t)^2$$

Answer 1

As suggested by @MartinR, consider the inequality $x^{2} \leq x^{2} + y^{2}$ and apply the squeeze theorem: \begin{align*} 0\leq \frac{x^{2}}{(x^{2} + y^{2})^{3/4}} \leq \frac{x^{2} + y^{2}}{(x^{2} + y^{2})^{3/4}} = (x^{2} + y^{2})^{1/4} \xrightarrow[]{(x,y)\to(0,0)} 0 \end{align*}

Hopefully this helps!