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.

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.
