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Prove that limit does not exist $$\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}$$ Obviously, since it is symmetric in $x$ and $y$, classic approach of substituing $x$ as a "simple" function - linear, polynomial, etc. Won't work. Polar gives us a reasonable answer $r\sin\theta\cos\theta$. How would I easily show it does not exist?

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Use $|xy|/\sqrt{x^2+y^2} \leq |y|$ to show that the limit exist. –  Winther Jun 23 at 1:14
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Sorry, it does exist, I thought of another one I've already proven, I should get some sleep. –  user74200 Jun 23 at 1:17
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@user74200 It happens to the best of us! –  Cameron Williams Jun 23 at 1:17

2 Answers 2

Using $$\frac{xy}{x^2+y^2}\leq \frac{1}{2}$$ we have that

$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leq \frac{1}{\sqrt{2}}\sqrt{xy}\rightarrow 0$$

so the limit exists and is zero.

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This doesn't quite make sense when $xy<0$. It's perhaps better (for more applications) to bound the numerator instead. –  Ted Shifrin Jun 23 at 4:02

As $(x,y)\to0$, we know $r\to0$, but $\sin\theta\cos\theta$ is bounded, so the limit does exist; it's zero.

One can plot the graph for verification.

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