Determine whether a function is continuous or differentiable

[trees and plants]

Homework Statement

Test at every point $(x_0, y_0) \in R^2$ this function

$f(x,y) =\frac{x^3y}{x^6 + y^2}$ if $(x,y)\neq (0, 0)$,

$0$ if $(x, y) = (0, 0)$, for continuity, partial differentiability, differentiability, continuous partial differentiability and compute (where it exists) its derivative.

Relevant Equations

$f(x,y) =\frac{x^3y}{x^6 + y^2}$ if $(x,y)\neq (0, 0)$,
$0$ if $(x, y) = (0, 0)$,

Perhaps use the definition of continuity, partial differentiability?

 

[trees and plants]

For the continuity of a function f we know that if $\forall (x_m)\subset U$ $(\forall\epsilon>0)(\exists \delta>0)$with $||x_m-x_0||< \delta \Rightarrow ||f(x_m)-f(x_0)||<\epsilon$,then the function is continuous at the point $(x_0,f(x_0))$ where $x_m$ is a sequence.

 

[trees and plants]

If we take $x_m=y_m=\frac{1}{m}$ then continuity still holds at $(0,0)$, what sequence should i choose to prove it does not hold at $(0,0)$?

 

[Mark44]

Homework Statement:: Test at every point $(x_0, y_0) \in R^2$ this function

$f(x,y) =\frac{x^3y}{x^6 + y^2}$ if $(x,y)\neq (0, 0)$,

$0$ if $(x, y) = (0, 0)$, for continuity, partial differentiability, differentiability, continuous partial differentiability and compute (where it exists) its derivative.
Relevant Equations:: $f(x,y) =\frac{x^3y}{x^6 + y^2}$ if $(x,y)\neq (0, 0)$,
$0$ if $(x, y) = (0, 0)$,

Perhaps use the definition of continuity, partial differentiability?

For f to be continuous at (0, 0), $\lim_{(x, y) \to (0, 0)} f(x, y)$ has to be 0 no matter which path is taken. I have found a path where the limit of this function is 1/2, which is enough to show that the function is not continuous at (0, 0). I leave it to you to figure out what path this is.

Your textbook should have some examples similar to this one.

 

[Delta2]

I think the path @Mark44 has in mind is $y=2x$.

 

[Mark44]

I think the path @Mark44 has in mind is y=2x.

Nope, that wasn't the one. It's not a straight line path.

 

[Delta2]

Yes i see now my mistake, its a higher degree curve right? better not to reveal it to the OP.

 

[Mark44]

Yes i see now my mistake, its a higher degree curve right? better not to reveal it to the OP.

Yes, higher degree.

 

[trees and plants]

Perhaps it is $y=x^3$?Is this correct?I do not understand what is the general case for this special case. If i had another example different from this, what should i do? What do you mean by path? Thank you.

 

[Mark44]

Perhaps it is $y=x^3$?Is this correct?

Yes. Along the curve $y = x^3$ the limit is 1/2 as (x, y) --> (0, 0). Along other curves, the limit is different. For example, along the curve $y = 2x^3$, the limit is 2/9 2/5 (corrected value).

I do not understand what is the general case for this special case. If i had another example different from this, what should i do?

There is no general case that I know of. You have to try different things for different problems.

What do you mean by path?

The curve along which the point (x, y) moves to the point in question, which in this problem is (0, 0).

 

[Stephen Tashi]

I do not understand what is the general case for this special case. If i had another example different from this, what should i do? What do you mean by path? Thank you.

The general technique is that if you suspect $lim_{(x,y) \rightarrow (a,b) } f(x,y)$ does not exist then try to find continuous functions $p_1(x)$ and $p_2(x)$ such that $lim_{x \rightarrow a} p_1(x) = b = lim_{x \rightarrow a} p_2(x)$ but $lim_{x\rightarrow a} f(x,p_1(x)) != lim_{x \rightarrow a} f(x,p_2(x))$.

In practical terms, look at the expression for $f(x,y)$ and see if you can make it take different values by enforcing different relations between the $x$ and $y$ variables. For example, suppose $f(x,y) = \frac {x + y}{x}$. Substituting $y = x$ gives a different result that substituting $y = 2x$. Both substitutions agree that $y = 0$ when $x = 0$.

You can consider the graphs $(x,p_1(x))$ and $(x, p_2(x))$ to be paths. This approach reduces the problem to working with limits involving only 1 variable. (Of course, sometimes you have to reverse the roles of $x$ and $y$)

The principle behind the general technique is the theorem that if $f(x,y)$ is continuous at $(a,b)$ as a function of 2 variables and $p(x)$ is any continuous function at $x = a$ such that $p(a) = b$ then $lim_{x \rightarrow a} f(x,p(x)) = f(a,b)$.

The contrapositive of this theorem says that if two functions such as the above $p_1$ and $p_2$ exist then $f(x,y)$ is not continuous at $(a,b)$.

It's an interesting question whether the converse of the above theorem holds. In other words, suppse we prove that there exists a number $L$ such that for any continuous path $(x,p(x)$ with $p(a) = b$ that $lim_{x \rightarrow a} f(x,p(x)) = L$. Can we then conclude that $lim_{(x,y) \rightarrow (a,b)} f(x,y) = L$?

 

[Office_Shredder]

The contrapositive as you stated is false for (imo) dumb reasons. Let f(x,y) equal 1 if x=0, 0 otherwise. You can't create a path under your set up because you eliminated the option of the vertical line as a path.

If you fully parameterize the path then it seems like a deep and interesting question (and I think the answer is yes it is continuous?)

 

[Delta2]

Something about differentiability and continuous partial differentiability: The way I was taught, if the partial derivatives of f exist and are continuous at a point, then we call the function f differentiable at that point.

So the notions of continuous partial differentiability and differentiability coincide. Is that correct?

 

[Eclair_de_XII]

The way I was taught, if the partial derivatives of f exist and are continuous at a point, then we call the function f differentiable at that point.

Do the values of each of the partials need to coincide at that point? I ask because I looked at this problem, and felt like doing it because I was rusty on my Calculus III, anyway. Anyway, I found a function $y=y(x)$ such that $f_Y(x,y)\neq f_X(x,y)$ at $(x,y)=(0,0)$, where $y\rightarrow 0$ as $x\rightarrow 0^+$. I mean, if we look at differentiability of a function of a single variable, the function cannot be differentiable at said point if the value of the derivative depends on the direction from which the point is approached. Think: $|x|$. Analogously, I think that a function of more than one variable cannot be differentiable at a point if again, the value of the derivative depends on the path taken to reach that point.

 

[Delta2]

Do the values of each of the partials need to coincide at that point? I ask because I looked at this problem, and felt like doing it because I was rusty on my Calculus III, anyway. Anyway, I found a function $y=y(x)$ such that $f_Y(x,y)\neq f_X(x,y)$ at $(x,y)=(0,0)$, where $y\rightarrow 0$ as $x\rightarrow 0^+$. I mean, if we look at differentiability of a function of a single variable, the function cannot be differentiable at said point if the value of the derivative depends on the direction from which the point is approached. Think: $|x|$. Analogously, I think that a function of more than one variable cannot be differentiable at a point if again, the value of the derivative depends on the path taken to reach that point.

Well from what i know, NO the partial derivatives need not be equal at that point, but need both to be continuous.
The definition of partial derivatives involves the limit of the rate of change as one variable approaches that point, so we have a single variable limit, so there are no path considerations except of course from the right and from the left.

 

[Eclair_de_XII]

If that's the case, then it should be somewhat easy to see why neither partial is continuous if the functions approach the origin along some trivial paths.

 

[Delta2]

Because the function is not continuous at (0,0) does this imply that the partial derivatives do not exist at (0,0)?

 

[Eclair_de_XII]

If I recall, if a function of one variable is differentiable, then it must be continuous. If it is not continuous, then the function cannot be differentiable. I'm still fuzzy on the details of partial derivatives and the derivative of functions of multiple variables. But if this function in particular is not continuous at the origin, then it should not be differentiable there, either, if we hold the previous statement for functions of one variable to be true for functions of multiple. As for how the non-differentiability of the function (if it is indeed not differentiable there) relates to the existence of partial derivatives at the origin... I do not recall.

 

[Eclair_de_XII]

I mean, if the function is not differentiable at the origin, then the graph of the function should not have a well-defined tangent plane at that point. And if there is something wrong with the tangent plane, then I can only assume that there is something wrong with the partial derivatives of the function, since the former depends on the latter. Maybe one of the partial derivatives is not well-defined or does not exist, is my guess.

 

[Delta2]

From what I know, if a multivariable function is not differentiable at a point then this means at least one of the following two things

• at least one of the partial derivatives does not exist at that point
• at least one of the partial derivatives is not continuous at that point

 

[jbunniii]

Because the function is not continuous at (0,0) does this imply that the partial derivatives do not exist at (0,0)?

No, both partial derivatives can exist at (0,0) even if the function is not continuous there. To see that this is possible, try evaluating your function on the x and y axes.

 

[MidgetDwarf]

If I recall, if a function of one variable is differentiable, then it must be continuous. If it is not continuous, then the function cannot be differentiable. I'm still fuzzy on the details of partial derivatives and the derivative of functions of multiple variables. But if this function in particular is not continuous at the origin, then it should not be differentiable there, either, if we hold the previous statement for functions of one variable to be true for functions of multiple. As for how the non-differentiability of the function (if it is indeed not differentiable there) relates to the existence of partial derivatives at the origin... I do not recall.

I think JBUNNiii is referring to C1 functions. If you want to read further on this subject. If I remember correctly, a differentiable function may not be C1.

 

[jbunniii]

I think JBUNNiii is referring to C1 functions. If you want to read further on this subject. If I remember correctly, a differentiable function may not be C1.

No, a $C^1$ function is a function which is differentiable and whose derivative is continuous.

The OP's function is not even continuous at $(0,0)$, hence is not differentiable at $(0,0)$, so it's certainly not $C^1$.

Despite the OP's function not being differentiable at $(0,0)$, both partial derivatives exist at $(0,0)$. (Since this is homework, I'll leave the proof of this to the OP.)

For a simpler function with the same properties, define $g(x,y) = 0$ on the $x$ and $y$ axes, and $g(x,y) = 1$ everywhere else.

 

[SammyS]

... Along other curves, the limit is different. For example, along the curve $y = 2x^3$, the limit is 2/9.

That 2/9 might be a typo.

It looks to me that the limit is 2/5 for that case

Along the curve $y=ax^3$, where $a$ is a constant, the limit as (x, y) --> (0, 0) is $\dfrac{a}{1+a^2}$.

 

[Mark44]

That 2/9 might be a typo.

It looks to me that the limit is 2/5 for that case

Yes, my mistake - thanks for pointing this out. I inadvertently cubed 2 instead of squaring it.
I have edited my earlier post.