Two-variable limit question $\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}$
How to solve this two-variable limit? Thanks :D
 A: Switching to polar coordinates:
$$
\lim_{r\to 0}\frac{r^4\cos^2\varphi\sin^2\varphi}{(r^2\cos^2\varphi+r^4\sin^4\varphi)r} = \lim_{r\to 0} \left[r\frac{\cos^2\varphi\sin^2\varphi}{\cos^2\varphi+r^2\sin^4\varphi}\right]
$$
then we have to be careful and see what happens as $\varphi$ changes its values: let us consider $\varphi=\pi /2 + k\pi$, then our quantity vanishes identically and the limit is zero; however if $\phi \ne \pi /2 + k\pi $, for $r>0$ the fraction is always bounded since it can be expressed as:
$$
\frac{\sin^2\varphi}{1+r^2\sin^2\varphi \tan^2\varphi},
$$
therefore the limits is again zero.
A: Hint : Put $x = r\cos(\theta)$ and $ y = r\sin(\theta)$ and change the limit to $r \to 0$ .
A: Let
\begin{equation}
f(x,y)=\frac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}
\end{equation}
If $y=0$, $\lim_{(x,y)\rightarrow(0,0)}f(x,y)=0$. If $y\neq0$, say approach the $(0,0)$ do not along with x-axis, we have:
\begin{equation}
f(x,y)=\frac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}\leq\frac{x^2y^2}{2|x|y^2\sqrt{x^2+y^2}}=\frac{|x|}{2\sqrt{x^2+y^2}}\\
0\leq\lim_{(x,y)\rightarrow(0,0)}f(x,y)\leq\lim_{(x,y)\rightarrow(0,0)}\frac{|x|}{2\sqrt{x^2+y^2}}=0
\end{equation}
In summary, $\lim_{(x,y)\rightarrow(0,0)}f(x,y)=0$.
A: This might also be an approach:
$$0 \le \frac{x^2y^2}{(x^2 + y^4)\sqrt{x^2 + y^2}} \le \frac{x^2y^2}{x^2\sqrt{x^2+y^2}} \le \frac{y^2}{\sqrt{y^2}} \le |y|$$
So, since $(x,y) \to (0,0)$ then $|y| \to 0$ and passing to a limit we obtain that the initial expression goes to 0.
