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$\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}$

How to solve this two-variable limit? Thanks :D

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    $\begingroup$ Welcome to Math.SE . What have you attempted till now ? $\endgroup$ – abkds Feb 17 '14 at 12:33
  • $\begingroup$ Did my hint help ? $\endgroup$ – abkds Feb 17 '14 at 12:47
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    $\begingroup$ Sure it helped a lot :) I didn't learn this method before. It is really useful :) And Brightsun confirmed my working. Thank you guys!! $\endgroup$ – SPMIHC Feb 17 '14 at 14:02
  • $\begingroup$ The model answer of this question is like this: $\dfrac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}} \leq \dfrac{x^2y^2}{(x^2+y^2)^{\frac{3}{2}}} < \sqrt{|xy|}$ Then it uses definition of limit to prove it equals 0. But I don't think the first part is true as $y$ can be less than 1. Is this proof correct? Thanks :) $\endgroup$ – SPMIHC Feb 17 '14 at 14:32
  • $\begingroup$ @SPMIHC you are welcome $\endgroup$ – Brightsun Feb 18 '14 at 12:56
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Switching to polar coordinates: $$ \lim_{r\to 0}\frac{r^4\cos^2\varphi\sin^2\varphi}{(r^2\cos^2\varphi+r^4\sin^4\varphi)r} = \lim_{r\to 0} \left[r\frac{\cos^2\varphi\sin^2\varphi}{\cos^2\varphi+r^2\sin^4\varphi}\right] $$ then we have to be careful and see what happens as $\varphi$ changes its values: let us consider $\varphi=\pi /2 + k\pi$, then our quantity vanishes identically and the limit is zero; however if $\phi \ne \pi /2 + k\pi $, for $r>0$ the fraction is always bounded since it can be expressed as: $$ \frac{\sin^2\varphi}{1+r^2\sin^2\varphi \tan^2\varphi}, $$ therefore the limits is again zero.

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    $\begingroup$ Yes, but then how you conclude that the limit is indeed 0? If the fraction was always bounded, then okay. But it is not the case. So what do you do? $\endgroup$ – Ant Feb 17 '14 at 13:13
  • $\begingroup$ @Ant I tried to explain what I had in mind. Is it correct now? $\endgroup$ – Brightsun Feb 18 '14 at 12:43
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    $\begingroup$ yes I think so! :) $\endgroup$ – Ant Feb 18 '14 at 12:49
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Hint : Put $x = r\cos(\theta)$ and $ y = r\sin(\theta)$ and change the limit to $r \to 0$ .

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Let \begin{equation} f(x,y)=\frac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}} \end{equation} If $y=0$, $\lim_{(x,y)\rightarrow(0,0)}f(x,y)=0$. If $y\neq0$, say approach the $(0,0)$ do not along with x-axis, we have: \begin{equation} f(x,y)=\frac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}\leq\frac{x^2y^2}{2|x|y^2\sqrt{x^2+y^2}}=\frac{|x|}{2\sqrt{x^2+y^2}}\\ 0\leq\lim_{(x,y)\rightarrow(0,0)}f(x,y)\leq\lim_{(x,y)\rightarrow(0,0)}\frac{|x|}{2\sqrt{x^2+y^2}}=0 \end{equation} In summary, $\lim_{(x,y)\rightarrow(0,0)}f(x,y)=0$.

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  • $\begingroup$ Thank you for your reply :) Could you explain more why $\lim_{(x,y)\rightarrow(0,0)}\frac{|x|}{2\sqrt{x^2+y^2}}$ equals 0? I am confused. Both the nominator and denominator tend to zero :O $\endgroup$ – SPMIHC Feb 17 '14 at 14:43
  • $\begingroup$ If we approach $(0,0)$ not along with x-axis, we can set $x=0$ first and obtain the limitation is $0$. And then no matter what value of $y$, the limitation is zero. $\endgroup$ – Lion Feb 17 '14 at 14:47
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This might also be an approach: $$0 \le \frac{x^2y^2}{(x^2 + y^4)\sqrt{x^2 + y^2}} \le \frac{x^2y^2}{x^2\sqrt{x^2+y^2}} \le \frac{y^2}{\sqrt{y^2}} \le |y|$$ So, since $(x,y) \to (0,0)$ then $|y| \to 0$ and passing to a limit we obtain that the initial expression goes to 0.

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