# Limit of a two variable function

• I
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?

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

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

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.

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

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

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.
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.

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

What are you taking the sin function of? X, y or some other function
Can you be more clear?

BvU
Homework Helper
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....

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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

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

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

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
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.

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....
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)

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BvU
Homework Helper
I'll flip into reading mode ... too many misses

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

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
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?

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Can I get back with you tomorrow on this? thanks

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

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

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.

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

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
It's a disk with radius approaching to 0. In case 0 <= t <= 2pi is independent and the limit become:

$$\lim_{r\rightarrow 0}\frac{r^3 \cos^3(t)+r^3 \sin^3(t)}{r\cos(t)+r^2\sin(t)}.$$
If t=pi/2 or 3pi/2 -> cos(t)=0 and sin(t)=+-1 and the limit converge to 0. Else we can do this:
$$\lim_{r\rightarrow 0}\frac{r^3 \cos^3(t)+r^3 \sin^3(t)}{r\cos(t)+r^2\sin(t)}=\lim_{r\rightarrow 0}\frac{r^2\cos^3(t)+r^2 \sin^3(t)}{\cos(t)+r\sin(t)}.=\frac{0}{\cos(t)}=0$$
Is this conclusive as a proof?

I would say no because this path may not work for other limit problems. Under the reasoning that if there was a "family" of paths that satisfied must multi variable limit problems it would have brought up when I took calculus courses years ago. What are your thoughts?

This limit does not exist. Note that the function isn't even defined on the path x=-y^2, and if you consider paths close to that path (e.g. x = -y^2 + y^4), the limit diverges to infinity.

This limit does not exist. Note that the function isn't even defined on the path x=-y^2, and if you consider paths close to that path (e.g. x = -y^2 + y^4), the limit diverges to infinity.
Which limit are you referring to?

Which limit are you referring to?
The one in the OP.

Stephen Tashi