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So I'm asked to prove $$\lim_{(x,y)->(0,0)}\frac{x^2y}{2x^2+y^2}$$ exists.

I've turned this into $$\lim_{(x,y)->(0,0)}\frac{x^2y}{2x^2+y^2}=\lim_{|p|->0}\frac{|p|^3}{2|p|^2}=0$$

Normally we have $|x|≤|p|$ and $|y|≤|p|$ but as $x,y->(0,0)$ these will be equal to each other, so I transformed to system in this way.

Do you think this is acceptable? I've never done such a thing before, Normally I would create an upper bound for the function and apply the squeeze theorem.

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  • $\begingroup$ This works for the numerator but doesn't work for the denominator. If $|x|\le |p|$ then $\frac{1}{|p|}\le \frac{1}{|x|}.$ $\endgroup$ – mfl Dec 3 '16 at 13:36
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No, that's not ok. $(x,y)\to (0,0)$ does not mean $|x|$ and $|y|$ will be equal. For example, it's possible to approach the origin along the line $y=2x$.

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  • $\begingroup$ What do you recommend then? $\endgroup$ – Xenidia Dec 3 '16 at 13:36
  • $\begingroup$ @Xenidia A squeeze argument should work if you note that $\dfrac{x^2}{2x^2+y^2} \le 1$ for all $(x,y) \ne (0,0)$ $\endgroup$ – tilper Dec 3 '16 at 13:47
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Note that $$\left|\frac{x^2y}{2x^2+y^2}-0\right|\leq \left|\frac{x^2y}{2x^2}\right|\leq\left|{y}\right|=\sqrt{y^2}\leq \sqrt{x^2+y^2}.$$

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