I tried following this answer (https://math.stackexchange.com/a/542303/86425), but I cannot see how this shows the limit is 0. Can someone please provide guidance
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Could you be a little more specific about what you don't understand in the answer? – Antonio Vargas Mar 12 '14 at 04:31
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I am confused as to the third term and onward, the term (sqrt(x^2+y^2)*sqrt(x^2+y^2))/sqrt(x^2+y^2) and onward. Why did we choose that expression? – user5826 Mar 12 '14 at 04:42
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Note that $\sqrt{x^2 + y^2} \geq \sqrt{y^2} = |y|$. Then $$ \left|\frac{xy}{\sqrt{x^2 + y^2}}\right| = |x|\frac{|y|}{\sqrt{x^2 + y^2}} \leq |x| $$ so $$ \lim_{(x,y) \to (0,0)}\left|\frac{xy}{\sqrt{x^2 + y^2}}\right| \leq \lim_{(x,y) \to (0,0)} |x| = 0 \, . $$
Personally, I prefer thinking of these problems in polar coordinates. So let $x = r \cos\theta$, $y = r \sin \theta$ where $r = \sqrt{x^2 + y^2}$. Then \begin{align*} \lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2 + y^2}} &= \lim_{r \to 0} \frac{r^2 \cos \theta \sin \theta}{r} = \lim_{r \to 0} r \cos \theta \sin \theta = 0 \, . \end{align*}
Viktor Vaughn
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I would've never thought about thinking in polar coordinates, how/where does that idea even come from? – user5826 Mar 12 '14 at 05:02
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It's a very standard technique for evaluating multivariable limits tending to the origin. – Viktor Vaughn Mar 12 '14 at 05:23
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$z=\frac{xy}{\sqrt{x^2+y^2}}\Longrightarrow 1/z^2=1/x^2+ 1/y^2\Longrightarrow\lim\limits_{(x,y)\to(0,0)} 1/z^2=\infty\Longrightarrow\lim\limits_{(x,y)\to(0,0)} z=0.$
Gil Bor
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