# I Limit of a two variable function

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1. Feb 23, 2017

### Drako Amorim

I'm trying to verify that: $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x+y^{2}}=0.$$
$$0<\sqrt{x^2+y^2}<\delta\rightarrow |\frac{\sin (x^{3}+y^{3})}{x+y^{2}}|<\epsilon$$
$$0\leq |\sin (x^{3}+y^{3})|\leq |(x^{3}+y^{3})|\leq |x|x^2+|y|y^2$$
$$|\frac{\sin (x^{3}+y^{3})}{x+y^{2}}|\leq \frac{ |x|x^2+|y|y^2}{|x+y^{2}|}$$
So I'm stuck here because of the denominator |x+y²|. What can I do?

2. Feb 24, 2017

### BvU

How about looking at $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})} { x^{3}+y^{3} } \ { x^{3}+y^{3} \over {x+y^{2}} } \rm ?$$

3. Feb 25, 2017

### Drako Amorim

Ok.
$$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x^{3}+y^{3}}\frac{x^{3}+y^{3})}{x+y^{2}}=\lim_{(x,y)\rightarrow (0,0)}\frac{x^{3}+y^{3}}{x+y^{2}}$$
The only progress that I have was reduce the limit to this:
$$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x^{3}+y^{3}}\frac{x^{3}+y^{3})}{x+y^{2}}=\lim_{(x,y)\rightarrow (0,0)} -\frac{xy}{x+y^2}$$

4. Feb 27, 2017

### Inventive

What do you get if you set x=0 and y be any value and take the limi? and do the same for y=0 and x any value and take the limit? Then try the limit of the function over the path x=y. In order for the limit to exist all of the selected paths must be continuous and have the same limit.

5. Feb 27, 2017

### BvU

Dint believe this is right. Can't follow how you go from $\ {x^{3}+y^{3}} \$ to $\ -xy\$ ? My idea for post #2 was: The first fraction gives you 1 in the limit, so all you have to worry about is $$\lim_{(x,y)\rightarrow (0,0)}\frac{x^{3}+y^{3}}{x+y^{2}}$$

6. Feb 27, 2017

### Drako Amorim

$$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x^{3}+y^{3}}\frac{x^{3}+y^{3}}{x+y^{2}}=\lim_{(x,y)\rightarrow (0,0)} \frac{x^{3}+y^{3}}{x+y^2}=\lim_{(x,y)\rightarrow (0,0)} \frac{(x+y^{2}-y^{2})x^{2}+(x+y^{2}-x)y}{x+y^2}=\lim_{(x,y)\rightarrow (0,0)} \frac{(x+y^2)x^2-x^2y^2+(x+y^2)y-xy}{x+y^2}=\lim_{(x,y)\rightarrow (0,0)} -\frac{x^2y^2+xy}{x+y^2}=\lim_{(x,y)\rightarrow (0,0)} -\frac{xy}{x+y^2}$$

All these 3 paths will give the result 0, but this does not prove that the lim will be 0 in another path, or I did not undertand your comentary.

7. Feb 27, 2017

### Inventive

What are you taking the sin function of? X, y or some other function

8. Feb 27, 2017

### Drako Amorim

Can you be more clear?

9. Feb 27, 2017

### BvU

What is $$\lim_{(x,y)\rightarrow (0,0)} \frac{(x+y^2)x^2} {x+y^2}\quad \rm ?$$

I see; probably not useful. But you are giving away far too much and end up with something that looks divergent for some $(x,y)\downarrow (0,0)$, so I don't believe your equals signs....

Last edited: Feb 27, 2017
10. Feb 27, 2017

### Inventive

Let's say x is turned off then you have f(y) = sin(y^3)/y^2 ? The limit (0,y) ->(0,0) is indeterminate at 0/0 and the limit is not 0. Likewise for (x,0) ->(0,0). Can you solve the limit it as functions of one variable for the function sin(y^3)/y^2? Hope this helps

11. Feb 27, 2017

### Inventive

I can see an operation on the function of f(x) or f(y) for the case of a indeterminate limit do you see it? And it works for g(x,y) = the sin function and h(x,y) =x+y^2. When I apply this rule my limit (x,y) -> (0,0) does equal 0. Could you explain why you want to prove this with the epsilon-delta method? It is going to be very difficult to show this as a proof. Hope this helps

12. Feb 27, 2017

### Drako Amorim

The limit is not 0? If you apply the L'hospital method it will give 0.

Can you show me what you're saying? You are saying that you can separate $$\frac{sin(x^3+y^3)}{x+y^2}$$ into $$f(x)g(y)$$?
About your question: This limit is a challenge that came from $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x^{2}+y^{2}}.$$ which is a lot more easy to verify.

I don't see any error in my calculations. If you trust in the Wolframalpha see these links:
https://www.wolframalpha.com/input/?i=lim+(x,y)->(0,0)+sin(x^3+y^3)/(x+y^2)
https://www.wolframalpha.com/input/?i=lim+(x,y)->(0,0)+(-xy)/(x+y^2)

Last edited: Feb 27, 2017
13. Feb 27, 2017

### BvU

I'll flip into reading mode ... too many misses

14. Feb 27, 2017

### Inventive

This is not multiplication of f(x) by g(y). It is f(y)/g(y). l'hopitals rule only applies to single variable limits of functions. I am trying to get you visually see how each path in a multi variable limit looks like one variable at a time. You will need to be creative in your path selections not just the curve in the yz plane (0,y) , xz plane (x,0) or even x= y to find the limit of 0 for your function

15. Feb 27, 2017

### Drako Amorim

ok. L'hospital rule was applied in this limit: $$\lim_{(0,y)\rightarrow (0,0)}\frac{\sin y^{3}}{y^{2}}.$$ edit: In fact this was a dumb way to solve this simple limit. You have a proof for this? Because paths do not are sufficient. Can you show me in algebra what you see?

Last edited: Feb 27, 2017
16. Feb 27, 2017

### Inventive

Can I get back with you tomorrow on this? thanks

17. Feb 28, 2017

### Inventive

Have you considered let x or y =t and the alternate variable equaling 0 and then taking the limit?

18. Feb 28, 2017

### Drako Amorim

Yes. x=t and y=0 so the limit will be:
$$\lim_{(t)\rightarrow (0)}\frac{\sin t^{3}}{t}=\lim_{(t)\rightarrow (0)}t^2\frac{\sin t^{3}}{t^3}=0.$$
x=0, y=t:
$$\lim_{(t)\rightarrow (0)}\frac{\sin t^{3}}{t^2}=\lim_{(t)\rightarrow (0)}t\frac{\sin t^{3}}{t^3}=0.$$
x=y=t:
$$\lim_{(t)\rightarrow (0)}\frac{\sin 2t^{3}}{t+t^2}=\lim_{(t)\rightarrow (0)}\frac{6t^2\cos 2t^{3}}{1+2t}=0.$$
Can I conclude something with x=rcos(t), y=rsin(t) doing r->0?

19. Feb 28, 2017

### Inventive

Interesting approach to do it from a polar equation point of view. Let me think about. Are we assuming a fixed value of t or is r and t variable. Based on what you wrote it looks like t is fixed. Maybe you can clarify that for me thanks.

20. Feb 28, 2017

### Inventive

I want to clarify my use of polar in terms of the equations that define a circle. I think you are describing a disk in the x,y plane approaching a radius of 0 or in three space a cone who radius is converging to 0