<s>[INST] Calculating $\lim_{(x,y) \to(0,0)} \frac{x^2y}{x^2+y^4}$ I can't figure out how to calculate this limit (or prove it does not exist)
$$
\lim_{(x,y) \to(0,0)} \frac{x^2y}{x^2+y^4}
$$
I've tried with restrictions on $y=mx$ and curves of the form $y=x^n$. 
The limit should not exist but even with polar coordinates I can't figure it out
 [/INST] Let $f(x,y)={\large{\frac{x^2y}{x^2+y^4}}}$.

Let $x^2+y^2=r^2$, with $0 < r \le 1$.

If $x\ne 0$, then
\begin{align*}
|f(x,y)|&=\left|\frac{x^2y}{x^2+y^4}\right|\\[4pt]
&\le\left|\frac{x^2y}{x^2+x^2y^4}\right|\;\;\;\;\;\text{[since $x^2\le r^2\le 1$]}\\[4pt]
&=\left|\frac{y}{1+y^4}\right|\\[4pt]
&\le |y|\\[4pt]
&\le r\\[4pt]
\end{align*}
and if $x=0$, then $y\ne 0$, so 
$$
f(x,y)=\frac{0}{y^4}=0
\qquad\qquad\qquad\qquad\qquad\;\;\;
$$

In either case, we have $|f(x,y)|\le r$.

Letting $r$ approach zero from above, it follows that 
$$
\lim_{(x,y)\to (0,0)}f(x,y)=0
\qquad\qquad\qquad\qquad\qquad\;\;\;
$$
</s>