I Limit of a two variable function

[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?

 

[BvU]

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

 

[Drako Amorim]

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}

 

[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.

 

[BvU]

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

 

[Drako Amorim]

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.

 

[Inventive]

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

 

[Drako Amorim]

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

Can you be more clear?

 

[BvU]

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

[edit]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...

 

[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

 

[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

 

[Drako Amorim]

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

[edit]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)

 

[BvU]

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

 

[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

 

[Drako Amorim]

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?

 

[Inventive]

Can I get back with you tomorrow on this? thanks

 

[Inventive]

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

 

[Drako Amorim]

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?

 

[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.

 

[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

 

[Drako Amorim]

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?

 

[Inventive]

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?

 

[Citan Uzuki]

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.

 

[Inventive]

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?

 

[Citan Uzuki]

Which limit are you referring to?

The one in the OP.

 

[Inventive]

I am not saying I agree with the answer, just that I have never tried that approach before. One problem with letting x=rcos(t) and y=rsin(t) is that it limits curve to travel at any point on the unit circle (assuming r=1 for arguments sake) for any value of t, which supports your comment about this problem's ability to have a limit that exists. Since we do not know where the problem came from (text book, etc). It could be that limit does not exist as stated and the sources answer is wrong to begin with. My next thought on this problem is to rewrite the numerator using identities and break it up into two separate limits and see where that goes.

 

[haushofer]

I'm trying to verify that: \lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x+y^{2}}=0.
0&lt;\sqrt{x^2+y^2}&lt;\delta\rightarrow |\frac{\sin (x^{3}+y^{3})}{x+y^{2}}|&lt;\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?

My 2 cents:

I'd evaluate the limit along a path like y=kx for real k; if the limit does not depend on k, it exists, and you can try to manage it in a form to apply the squeeze theorem. So, evaluating the limit along the path y(x)=kx gives
\lim_{ x\rightarrow 0}\frac{\sin ([k^3+1] x^3)}{x+k^2 x^2}

Now use the inequality
-t \leq \sin(t) \leq +t

for real t to constrain the limit by

- \frac{[k^3+1] x^3}{x+k^2 x^2} \leq \lim_{ x\rightarrow 0}\frac{\sin ([k^3+1] x^3)}{x+k^2 x^2} \leq \frac{[k^3+1] x^3}{x+k^2 x^2}

Using L'Hospital on the left and right gives

\lim_{x\rightarrow 0} \pm \frac{[k^3+1] x^3}{x+k^2 x^2} = \lim_{x\rightarrow 0} \pm \frac{[k^3+1] 3x^2}{1+2k^2 x} = \pm \frac{0}{1+0} = 0

so according to the squeeze theorem one has

\lim_{ x\rightarrow 0}\frac{\sin ([k^3+1] x^3)}{x+k^2 x^2} =0

Since this is true for all k, the answer depends not on the angle to which one arrives at the point (x,y)=(0,0), so the limit exist
and is equal to

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

 

[Inventive]

That makes complete sense when using the squeeze theorem for the path y(x) =kx. Earlier in the posts where y=x was tried it, would have worked then if the squeeze theorem was applied at that point in time to the problem. I forgot all about suggesting that theorem be tried again. Thank you!

 

[haushofer]

Trying the path y(x)=x can only indicate that the limit does not exist if it is supplemented by another path giving different answers. That's the nice thing about paths like y(x)=kx or switching to polar coordinates or other curve representations that cover all the directs in which one approaches the limit point (x,y); they are general.

 

[Stephen Tashi]

Since this is true for all k, the answer depends not on the angle to which one arrives at the point (x,y)=(0,0), so the limit exist

That is not a valid conclusion. The existence of a common value for the limit along a family of straight line paths does not imply that the limit as (x,y) -> (0,0) exists. For example, see the answer to http://math.stackexchange.com/quest...existence-of-limit-of-a-two-variable-function

 

[Stephen Tashi]

This limit does not exist. Note that the function isn't even defined on the path x=-y^2,

That's correct, but does that, by itself, prevent the limit from existing? This is very technical question involving how we interpret the definition of the limit. The definition has the form " For each ... there exists...such that...for each $(x,y)$, $| f(x,y) - L| < \epsilon$". $\$ Are we to understand that the implicit meaning is "for each $(x,y)$ in the domain of $f$"? $\$ Or perhaps that $| f(x,y) - L | < \epsilon$" is not a false statement when $(x,y)$ is not in the domain of $f$, but rather an undefined expression?

and if you consider paths close to that path (e.g. x = -y^2 + y^4), the limit diverges to infinity.

That's a good idea, but does it work? $\ | sin(\theta) | \le |\theta |$. $\ \theta$ would be a polynomial of degree 12 in $y$ and the denominator would be a polynomial of degree 4 in $y$.

 

[Stephen Tashi]

I'm trying to verify that: \lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x+y^{2}}=0.

One thought is that $|\sin(\theta)| \le |\theta|$. $\$ Argue that for $(x,y) \ne (0,0)$, $$ | \frac{ \sin(x^3 + y^3)}{x + y^2} | \le |\frac{x^3 + y^3}{x + y^2}|$$. Then try polar coordinates on the ratio of the polynomials.

\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}

That's a good intuitive idea, but notice how hard it is to make that algebra into correct logic. For example, it assumes at the outset that $lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x^{3}+y^{3}}\frac{x^{3}+y^{3}}{x+y^{2}}$ exists. Of course, some instructors (and many home work graders) prefer to see algebra instead of logic.

The basic pattern that obeys logic would be:
Show $lim_{(x,y) \rightarrow (0,0)} p(x,y) = L_1$
Show $lim_{(xy) \rightarrow (0,0)} q(x,y) = L_2$
Conclude that $lim_{(x,y) \rightarrow (0,0)} p(x,y)q(x,y) = L_1 L_2$.

Polar coordinates might be a good way to tackle $\lim_{(x,y)\rightarrow (0,0)} \frac{x^{3}+y^{3}}{x+y^2}$ instead of using further algebra with $x$'s and $y$'s.

 

[Inventive]

How about this approach. I plotted this function using quick graph. There are no breaks in the graph at the origin so therefore it is continuous there. Also the numerator sin(x^3 + y^3) can be rewritten as sin(x^3)cos(y^3) + cos(x^3)sin(y^3). Now that the numerator has been rewritten as a sum of the products of periodic continuous functions it should make it easier to prove the existence of the limit. Even though the denominator is undefined at (0,0), the numerator is approaching 0 faster than the denominator in terms of how the graph is changing.

 

[Stephen Tashi]

There are no breaks in the graph at the origin so therefore it is continuous there.

The function $\frac{\sin(x^3 + y^3)}{x + y^2}$ is not continuous at $(0,0)$. Have you studied the definition of "$f(x,y)$ is continuous at $(a,b)$" ?

 

[Inventive]

i don't understand the "*. *". Comment. This site is not about PROVING someone is wrong or putting people down which you did with me. The dialog in these forums is not supposed to be judgmental towards others. So tell me why you disagree without being judgmental. What do you see in detail is wrong with my approach ?

 

[Inventive]

See my note

 

[Citan Uzuki]

That's correct, but does that, by itself, prevent the limit from existing? This is very technical question involving how we interpret the definition of the limit. The definition has the form " For each ... there exists...such that...for each $(x,y)$, $| f(x,y) - L| < \epsilon$". $\$ Are we to understand that the implicit meaning is "for each $(x,y)$ in the domain of $f$"? $\$ Or perhaps that $| f(x,y) - L | < \epsilon$" is not a false statement when $(x,y)$ is not in the domain of $f$, but rather an undefined expression?

All very true, but actually orthogonal to the point I was trying to make. What I was trying to get at is that if you are trying to find the limit of a rational function (or something which is effectively a rational function, since \sin (\theta) ~ \theta for small \theta) at the origin, the denominator being zero along some curve running through the origin usually means that you can make the function blow up by following another curve close to the first one. Thus seeing that the function fails to exist along some curve through the origin is usually a good sign that the limit won't exist, independently of the answer to that technical question.

That's a good idea, but does it work? $\ | sin(\theta) | \le |\theta |$. $\ \theta$ would be a polynomial of degree 12 in $y$ and the denominator would be a polynomial of degree 4 in $y$.

Yes. Along the curve x=-y^2 + y^4:

\lim_{y \to 0^+} \frac{\sin (x^3 + y^3)}{x + y^2} \\<br /> = \lim_{y \to 0^+}\frac{\sin ((-y^2 + y^4)^3 + y^3)}{-y^2 + y^4 + y^2} \\<br /> = \lim_{y \to 0^+}\frac{\sin (y^3 - y^6 + 3y^8 - 3y^{10} + y^{12})}{y^4} \\<br /> = \lim_{y \to 0^+}\frac{y^3 + O(y^6)}{y^4} \\<br /> = \lim_{y \to 0^+}\frac{1 + O(y^3)}{y} \\<br /> = \infty

So the original limit does not exist.

 

[Inventive]

All very true, but actually orthogonal to the point I was trying to make. What I was trying to get at is that if you are trying to find the limit of a rational function (or something which is effectively a rational function, since \sin (\theta) ~ \theta for small \theta) at the origin, the denominator being zero along some curve running through the origin usually means that you can make the function blow up by following another curve close to the first one. Thus seeing that the function fails to exist along some curve through the origin is usually a good sign that the limit won't exist, independently of the answer to that technical question.
Yes. Along the curve x=-y^2 + y^4:

\lim_{y \to 0^+} \frac{\sin (x^3 + y^3)}{x + y^2} \\<br /> = \lim_{y \to 0^+}\frac{\sin ((-y^2 + y^4)^3 + y^3)}{-y^2 + y^4 + y^2} \\<br /> = \lim_{y \to 0^+}\frac{\sin (y^3 - y^6 + 3y^8 - 3y^{10} + y^{12})}{y^4} \\<br /> = \lim_{y \to 0^+}\frac{y^3 + O(y^6)}{y^4} \\<br /> = \lim_{y \to 0^+}\frac{1 + O(y^3)}{y} \\<br /> = \infty

So the original limit does not exist.

Thanks for taking the time to explain it this way. I can visualize what you are saying

 

[Stephen Tashi]

Yes. Along the curve x=-y^2 + y^4:

I see what you mean.

We can also use the way more often seen in textbooks - showing that different limits can be had by using different curves if we use curves of the form $x = -y^2 + ky^3$.

Let $G(x,y) = \frac{x^3 + y^3}{x + y^2}\$ Use the path defined by $x = -y^2 + k y^3$ where $k$ is a non-zero constant. This path goes through $(0,0)$
Along this path, for $y \ne 0$, $G(x,y) = g(y) = \frac{ (-y^2 + ky^3)^3 + y^3}{ky^3} = \frac{(y^6 + 3ky^7 - 3k^2 y^8 + k^3 y^9 ) + y^3}{ky^3} = \frac{y^3+ 3ky^4 - 3k^2y^5 + y^6 + 1}{k}$
So $lim_{y \rightarrow 0} g(y) = \frac{1}{k}$.
Since $\sin()$ is a continuous function $lim_{y \rightarrow 0} \sin(g(y)) = \sin( \frac{1}{k})$.
This shows $lim_{(x,y) \rightarrow (0,0)} sin(G(x,y))$ does not exist since we have can get different limits on different paths depending on our choice for $k$.

 

[Inventive]

Thank you Stephen for the further information

 

[haushofer]

That is not a valid conclusion. The existence of a common value for the limit along a family of straight line paths does not imply that the limit as (x,y) -> (0,0) exists. For example, see the answer to http://math.stackexchange.com/quest...existence-of-limit-of-a-two-variable-function

You're right. The first example of your link is even used in stewart's textbook. Thanks for noting!