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Prove that $$\lim_{(x,y)\to(0,0)} \frac{x^3y}{x^6+y^2} = 0.$$

The only why I can think about is using the Sandwich theorem.

Because $\lim_{(x,y)\to(0,0)} \frac{x^3y}{x^6+y^2} = 0$, then I just need to find $h(x,y)$ such that $\lim_{(x,y)\to(0,0)} h(x,y) = 0$ such that:

$$0 \le \frac{x^3y}{x^6+y^2} \le h(x).$$

How can I find $h(x)$?

EDIT:

So what is wrong with my wolframalpha query? why is it says the limit is zero?

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  • $\begingroup$ Is $\frac{x^{3}y}{x^{6}+y^{2}}$ always non-negative? Even for points arbitrarily close to the origin? $\endgroup$ – T. Eskin Dec 6 '13 at 3:16
  • $\begingroup$ @ThomasE. What do you mean? $\endgroup$ – Billie Dec 6 '13 at 3:17
  • $\begingroup$ @user1798362 : You have written $0 \le \frac{x^3y}{x^6+y^2} \le h(x)$.. Mr.Thomas wants to know if $\frac{x^3y}{x^6+y^2}$ is always positive.. what does this equal to at $(-1,1)$ $\endgroup$ – user87543 Dec 6 '13 at 3:26
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    $\begingroup$ @user1798362: What if $x^3 = y$? $\endgroup$ – user99914 Dec 6 '13 at 3:27
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This limit is not equal to $0$ . When the limit exists , if you come by any path to that point the limit should be the same and it should be finite .

You can verify that if I go to the point $(0,0)$ by using $ y = x $ the limit tends to $0$ but when I use $y = x^3$ , the limit comes to be $0.5$ . Hence the limit does not exist

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  • $\begingroup$ wolframalpha.com/input/… So why is wolframAlpha says the limit is 0? $\endgroup$ – Billie Dec 6 '13 at 3:35
  • $\begingroup$ I think it depends on what type of algorithm they are using to find the limit . Any algorithm can't cover all the paths ( I think so ) that is why the erroneous limit . $\endgroup$ – abkds Dec 6 '13 at 3:40
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For $ y=mx^3$,

$$ \frac{x^3y}{x^6 + y^2} =\frac{m}{1+m^2} $$

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  • $\begingroup$ wolframalpha.com/input/… So why is wolframAlpha says the limit is 0? $\endgroup$ – Billie Dec 6 '13 at 3:34
  • $\begingroup$ @user1798362: wolframalpha calculate wrong! This limit depend on the variable m, thus limit does not exist $\endgroup$ – Iloveyou Dec 7 '13 at 5:54
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In particular if you use $y=mx^3$ and proceed to evaluate the limit along this path then

$\lim_{(x,mx^3)\to(0,0)} \frac{x^3y}{x^6+y^2}=\frac {m}{1+m^2}$ and for different values of $m$ you will have different limit values

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