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$$\lim_{(x,y) \to (0,0)} \frac{x y^2}{(3x^2 + 4x^2)}$$

How would one calculate above "using polar coordinates"? It was mentioned during class shortly, but we won't be introduced to this until next semester, yet I'd like to know.

It's a "$0/0$" expression, but I didn't use the method where I look at different "paths", as the teacher said the limit exists, thus using the path method wouldn't get me anywhere.

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    $\begingroup$ You might have a typo in the denominator. If the denominator is supposed to be $3x^2+4y^2$ and if you care, you don't need to use polar coordinates to easily find the limit. $\endgroup$
    – Git Gud
    Oct 26 '14 at 14:35
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Hints:

So you can suppose that $x=r\cos \theta$ and $y=r\sin\theta$ and notice that $r \to 0$ as $(x,y) \to (0,0)$

$$\lim_{(x,y) \to (0,0)} \frac{x y^2}{(3x^2 + 4x^2)}=\lim_{r\to 0}\frac{r^3\cos\theta\sin^2\theta}{3r^2\cos^2\theta+4r^2\sin^2\theta}=0$$

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One way is to use elliptic coordinates: write $x = \dfrac{1}{\sqrt{3}} r\cos \theta$ and $y = \dfrac{1}{2} r\sin \theta$. (Which makes for a much shorter computation than polar coordinates.)

Then (assuming the denominator is supposed to be $3x^2 + 4y^2$): $$ \frac{xy^2}{3x^2 + 4y^2} = \frac{\frac16 r^3 \cos \theta \sin^2 \theta}{r^2} = \frac16 r \cos\theta \sin^2 \theta. $$ When $(x,y) \to (0,0)$, $r \to 0$, and $$ \left|\frac{xy^2}{3x^2 + 4y^2}\right| = \frac16 r |\cos\theta \sin^2 \theta| \le \frac16 r $$ which tends to $0$ as $r \to 0$.

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  • $\begingroup$ how do you know that $r|\cos(\theta)\sin^2(\theta)|\leq r$ for every value of $\theta?$ What if $\cos(\theta)\sin^2(\theta)=1/r$? $\endgroup$
    – michek
    Mar 27 '15 at 14:50
  • $\begingroup$ @michek: $|\cos \theta| \le 1$ and $|\sin \theta| \le 1$ for all $\theta$. $\endgroup$
    – mrf
    Mar 27 '15 at 14:55
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It's all about making the change of variables:

$$x = \rho \cos{t}, \quad y = \rho \sin{t}, \quad t \in [0,2\pi), \quad \rho \geq 0. $$

In your case, as $(x,y)$ approaches the origin, the radius $\rho$ goes to $0$ for any angle $t$. Can you take it from here?

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Let $x=r\cos(\theta),y=r\sin(\theta)$, and let $r,\theta$ be continuous functions of some parameter $t$, such that $r(0)=0$ and $r(t)\geq 0$ for all $t$.

Then we have $$\frac{xy^2}{(3x^2+4y^2)}=\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2(3\cos^2(\theta)+4\sin^2(\theta))}=r\frac{\cos(\theta)\sin^2(\theta)}{(3+\sin^2(\theta))}$$ Furthermore, $$\left\vert\frac{xy^2}{(3x^2+4y^2)}\right\vert=\left\vert r\frac{\cos(\theta)\sin^2(\theta)}{(3+\sin^2(\theta))}\right\vert\leq \frac{r}{3}\to0\text{ as }r\to0.$$

Taking any continuous path of $(x,y)$ to zero is the same as letting $t\to0$ for appropriate $r(t),\theta(t)$. Since $r(t)\to0$ as $t\to0$, we get $$\frac{xy^2}{(3x^2+4y^2)}\to 0\text{ as }(x,y)\to0\text{, regardless of the path of approach.}$$

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  • $\begingroup$ (Assuming that the question was mistyped.) $\endgroup$ Oct 26 '14 at 14:51

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