Limit of $\lim_{(x,y)\to (0,0)}\frac{3x^3y^2+xy^4}{(x^2+y^2)^2}$ I've run across a particular limit in my multivariable calculus class, that being:
\begin{equation}
\lim_{(x,y)\to (0,0)}\frac{3x^3y^2+xy^4}{(x^2+y^2)^2}.
\end{equation}
I'm having quite a bit of trouble finding this limit using the squeeze theorem. Using polar coordinates (and assuming i'm not completely butchering everything), I believe we have that this particular limit evaluates to 0; however, using coordinate changes with limits feels a little uncomfortable and I would like to see if it can be done without it.
 A: We have
$$0\le \left|\frac{3x^3y^2+xy^4}{(x^2+y^2)^2}\right|\le \frac{3|x^3|y^2+|x|y^4}{(x^2+y^2)^2}\le$$
$$\le \frac{3|x|x^2y^2+\frac32|x|y^4}{2x^2y^2+y^4}=\frac32|x|\to0$$
A: You can use that
$$\left\lvert\frac{3x^3y^2+xy^4}{(x^2+y^2)^2}\right\rvert\leq\frac{3\lvert x\rvert(x^2+y^2)^2+\lvert x\rvert(x^2+y^2)^2}{(x^2+y^2)^2}=4\lvert x\rvert\to0,$$
as $(x,y)\to(0,0)$, which I got from just using the triangle inequality and that
$$x^2\leq x^2+y^2,\quad y^2\leq x^2+y^2.$$
A: Using polar coordinates is perfectly alright and does not require any special knowledge.  The formal argument can be presented as follows:
Let $\varepsilon > 0$.  Take $\delta = \varepsilon / 4$ and assume $|(x,y)| =  (x^2+y^2)^{1/2} < \delta$.  We can always find $r, \theta$ such that $x = r\cos\theta $ and $y = r\sin\theta$.  Then $r^2 = x^2 + y^2$ and our choice of $(x,y)$ implies $r < \delta$.
The numerator satisfies,
\begin{align}
| 3 r^5 \cos^3 \theta \sin^2 \theta + r^5 \cos\theta \sin^4\theta| &= r^5\lvert 3cos^3\theta \sin^2\theta + \cos\theta\sin^4\theta\rvert \\
&\leqslant 4r^5 \\
&<\varepsilon r^4
\end{align}
and the denominator is simply $r^4$, so that,
\begin{align}
\left\lvert \frac{3x^2y^3+xy^4}{(x^2+y^2)^2}\right\rvert < \varepsilon.
\end{align}
Since $\varepsilon$ is arbitrary, the required limit follows.
