Find limit of $\frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2}$ I have to show that $\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2} = 0$. But I cannot figure out the trick you need to find an upper estimation which goes to $0$. Do you have any hints?
EDIT:
I think I got it:
$\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2} =  \lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2}{(x^4+2x^2 y^2 +y^4)} +\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|y^4}{(x^4+2x^2 y^2 +y^4)}\leq \lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|^3 y^2}{x^2 y^2}+\lim_{{x\choose y}\to { 0 \choose 0}} \frac{|x|y^4}{y^4}=\lim_{{x\choose y}\to { 0 \choose 0}} |x|+ \lim_{{x\choose y}\to { 0 \choose 0}} |x|=0.$
What do you think?

Your solution is fine indeed
$$\frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2} \le\frac{|x|^3 y^2}{x^2y^2}+\frac{|x|y^4}{y^4}=2|x| \to 0$$
as an alternative by polar coordinates we have
$$\frac{|x|^3 y^2+|x|y^4}{(x^2+y^2)^2}=r \cdot f(\theta) \to 0$$
since $f(\theta)$ is bounded.
