Limit of multivariare function at the origin Consider the function $f:\mathbb{R}^2\to \mathbb{R}$ defined as:
$$
f(x,y)=
\begin{cases}
\frac{|x|^{\frac{5}{2}}y}{(x^2+y^4)\sqrt{x^2+y^2}}.\quad&\text{ if } (x,y)\neq 0\\
0\quad& \text{ if }(x,y)= 0\,.
\end{cases}
$$
Is this function continuous at the origin?
If the limit exists it has to be $0$ since, for example, if we take the restriction $x=y$ we obtain:
$$
\lim_{x\to0}\frac{|x|^{\frac{5}{2}}x}{\sqrt{2}(x^2+x^4)|x|}=0
$$
Restrictions to any kind of powers seem to give the same result suggesting that the function has to be continuous at the origin (as the graph also seems to confirm) but I'm not able to find an estimate for $f$ to use the squeeze theorem and actually prove continuity.
Any help will be greatly appreciated.
 A: We have
$$
\frac{|x|^{\frac{5}{2}}y}{(x^2+y^4)\sqrt{x^2+y^2}}=
|x|^{\frac{1}{2}}\cdot\frac{|x|^{2}}{x^2+y^4}\cdot
\frac{y}{\sqrt{x^2+y^2}}.
$$
Since
$$
\left|\frac{|x|^{2}}{x^2+y^4}\right|=\frac{x^{2}}{x^2+y^4}\leq 1
$$
and
$$
\left|\frac{y}{\sqrt{x^2+y^2}}\right|=\frac{|y|}{\sqrt{x^2+y^2}}\le1
$$
we obtain
$$
\left| \frac{|x|^{\frac{5}{2}}y}{(x^2+y^4)\sqrt{x^2+y^2}} \right| \leq |x|^{\frac{1}{2}}
$$
or
$$
-|x|^{\frac{1}{2}}\leq \frac{|x|^{\frac{5}{2}}y}{(x^2+y^4)\sqrt{x^2+y^2}} \leq |x|^{\frac{1}{2}}.
$$
A: Alternatively:
$$\sqrt{\frac{x^2+y^2}2}\ge \sqrt{|xy|}\implies \frac{\sqrt{2}}{\sqrt{x^2+y^2}}\le\frac{1}{\sqrt{|xy|}}\tag 1$$
so $$\frac{|x|^{5/2}|y|}{(x^2+y^4)\sqrt{x^2+y^2}}=\frac{x^2\sqrt{|xy|}\sqrt{|y|}}{\sqrt 2(x^2+y^4)}\frac{\sqrt 2}{\sqrt{x^2+y^2}}\le\sqrt{|y|}\frac{x^2\sqrt{|xy|}}{\sqrt 2 x^2\sqrt{|xy|}}\le\sqrt{|y|}$$
or from $(1)$, you could've immediately written $\frac{\sqrt{|xy|}}{\sqrt{x^2+y^2}}\le\frac1{\sqrt 2}.$
A: Yet another estimate:
$$
\dfrac{|x|^{5/2}|y|}{(x^2+y^4)\sqrt{x^2+y^2}}\leq \dfrac{\left(\sqrt{x^2}\right)^{5/2}\sqrt{x^2+y^2}}{(x^2+y^4)\sqrt{x^2+y^2}} \leq \dfrac{(x^2+y^4)^{5/4}}{x^2+y^4} = (x^2+y^4)^{1/4}\to 0 \quad(x,y \to 0)
$$
