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

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?

• Is $\frac{x^{3}y}{x^{6}+y^{2}}$ always non-negative? Even for points arbitrarily close to the origin? – T. Eskin Dec 6 '13 at 3:16
• @ThomasE. What do you mean? – Billie Dec 6 '13 at 3:17
• @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 $(-1,1)$ – user87543 Dec 6 '13 at 3:26
• @user1798362: What if $x^3 = y$? – user99914 Dec 6 '13 at 3:27

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

• wolframalpha.com/input/… So why is wolframAlpha says the limit is 0? – Billie Dec 6 '13 at 3:35
• 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 . – abkds Dec 6 '13 at 3:40

For $y=mx^3$,

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

• wolframalpha.com/input/… So why is wolframAlpha says the limit is 0? – Billie Dec 6 '13 at 3:34
• @user1798362: wolframalpha calculate wrong! This limit depend on the variable m, thus limit does not exist – Iloveyou Dec 7 '13 at 5:54

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