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How can I calculate the multivariable limit of $\lim_{(x,y) \to (0,0)} \frac{2xy^4}{(x^2+y^2)^2}$?

I'm new to this and I've seen a couple of examples, where it is possible to limit the fraction by a another fraction with only one variable. I tried this idea but it does not lead me anywhere.

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    $\begingroup$ Use polar coordinates for that. $\endgroup$ – Bernard May 21 at 19:50
  • $\begingroup$ Chiming in with Bernard. Whenever $x^2+y^2$ appears in a role, the use of polar coordinates is somewhat indicated. There are exceptions, but they are then high on the list of things to try. $\endgroup$ – Jyrki Lahtonen May 21 at 19:53
  • $\begingroup$ Thanks for your advice. $\lim_{(x,y) \to (0,0)} \frac{2xy^4}{(x^2+y^2)^2} = \lim_{r \to 0} \frac{2 r cos \theta r^4 sin\theta}{r^4} = \lim_{r \to 0} 2 r cos \theta sin\theta = 0$ Is this correct? Why can I just assume $r \to 0$? Wouldn't that imply that I am only regarding "straight lines" to zero? What about all the other potential $(x,y) \to (0,0)$? $\endgroup$ – thanc May 21 at 20:22
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With polar coordinates this is a rather easy exercise, and without them you can try:

$$\left|\frac{2xy^4}{(x^2+y^2)^2}\right|\le2|x|\frac{y^4}{y^4}=2|x|\xrightarrow[(x,y)\to(0,0)]{}0$$

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