# Continuity of $g(x,y)$ and its partials

1. Homework Statement

Let $g(x,y)=\sqrt[3]{xy}$

a) Is $g$ continuous at $(0,0)$?

b) Calculate $\frac {\partial g}{\partial x}$ and $\frac{\partial g}{\partial y}$ when $xy \neq 0$

c) Show that $g_{x}(0,0)$ and $g_{y}(0,0)$ exist by supplying values for them.

d) Are $\frac {\partial g}{\partial x}$ and $\frac {\partial g}{\partial y}$ continuous at $(0,0)$?

e) Does the graph of $z=g(x,y)$ have a tangent plane at $(0,0)$? You might consider creating a graph of this surface.

f) Is $g$ differentiable at $(0,0)$?

2. Homework Equations

3. The Attempt at a Solution
a) $$lim_{y \rightarrow 0 \space along\space x=0}\space g(x,y) =0$$
$$lim_{x \rightarrow 0 \space along\space y=0}\space g(x,y)=0$$
$$lim_{y=mx\rightarrow 0} = 0$$

Therefore limit exists and the limit is $0$. Since $$g(0,0)=0$$ and is equal to limit, $g$ is continuous at $(0,0)$.

b) $$\frac {\partial g}{\partial x}=\frac{\frac{1}{3}y^\frac{1}{3}}{x^\frac{2}{3}}$$

$$\frac{\partial g}{\partial y}=\frac{\frac{1}{3}x^\frac{1}{3}}{y^\frac{2}{3}}$$

c) I'm not sure what they meant by supplying values for them. But, what I did was giving the value of zero to both of the partials. I'm not sure how to do this part of the question.

d) For $$\frac {\partial g}{\partial x}$$, $$lim_{x \rightarrow 0\space along\space y=0} \frac{\partial g}{\partial x}$$
$$= \frac{0}{x^\frac{2}{3}} = 0$$
$$lim_{y \rightarrow 0\space along\space x=0} \frac{\partial g}{\partial x}$$ is undefined.

Therefore, $g_{x}$ is not continuous at $(0,0)$. The same result is obtained for $g_{y}$.

e) I don't know. Although in (b), it is shown that $g_{x}$ and $g_{y}$ exist at $(0,0)$, since the partials are not continuous, I would say that there is no tangent plane at the origin.

f) Yes, I would say the function is differentiable at origin, because I can differentiate it using the chain rule. What has differentiability got to do with continuity? Why is this question asked? Does differentiability imply continuity?

It is a lot of questions, but I hope someone can help me out here.

Thanks.

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Ray Vickson
Homework Helper
Dearly Missed
1. Homework Statement

Let $g(x,y)=\sqrt[3]{xy}$

a) Is $g$ continuous at $(0,0)$?

b) Calculate $\frac {\partial g}{\partial x}$ and $\frac{\partial g}{\partial y}$ when $xy \neq 0$

c) Show that $g_{x}(0,0)$ and $g_{y}(0,0)$ exist by supplying values for them.

d) Are $\frac {\partial g}{\partial x}$ and $\frac {\partial g}{\partial y}$ continuous at $(0,0)$?

e) Does the graph of $z=g(x,y)$ have a tangent plane at $(0,0)$? You might consider creating a graph of this surface.

f) Is $g$ differentiable at $(0,0)$?

2. Homework Equations

3. The Attempt at a Solution
a) $$lim_{y \rightarrow 0 \space along\space x=0}\space g(x,y) =0$$

$$lim_{x \rightarrow 0 \space along\space y=0}\space g(x,y)=0$$

$$lim_{y=mx\rightarrow 0} = 0$$

Therefore limit exists and the limit is $0$. Since $$g(0,0)=0$$ and is equal to limit, $g$ is continuous at $(0,0)$.

b) $$\frac {\partial g}{\partial x}=\frac{\frac{1}{3}y^\frac{1}{3}}{x^\frac{2}{3}}$$

$$\frac{\partial g}{\partial y}=\frac{\frac{1}{3}x^\frac{1}{3}}{y^\frac{2}{3}}$$

c) I'm not sure what they meant by supplying values for them. But, what I did was giving the value of zero to both of the partials. I'm not sure how to do this part of the question.

d) For $$\frac {\partial g}{\partial x}$$, $$lim_{x \rightarrow 0\space along\space y=0} \frac{\partial g}{\partial x}$$

$$= \frac{0}{x^\frac{2}{3}}$$

$$=0$$

$$lim_{y \rightarrow 0\space along\space x=0} \frac{\partial g}{\partial x}$$ is undefined.

Therefore, $g_{x}$ is not continuous at $(0,0)$. The same result is obtained for $g_{y}$.

e) I don't know. Although in (b), it is shown that $g_{x}$ and $g_{y}$ exist at $(0,0)$, since the partials are not continuous, I would say that there is no tangent plane at the origin.

f) Yes, I would say the function is differentiable at origin, because I can differentiate it using the chain rule. What has differentiability got to do with continuity? Why is this question asked? Does differentiability imply continuity?

It is a lot of questions, but I hope someone can help me out here.

Thanks.
You cannot assume that $\lim_{x,y \to 0} g_x(x,y)$ is $g_x(0,0)$. The actual definition of $g_x(0,0)$ is
$$g_x(0,0) = \lim_{h \to 0} \frac{g(h,0) - g(0,0)}{h}$$
if the limit exists.

For multivariate functions, what is the actual definition of the term "differentiable"? Be warned: it may not be the same as for a single-variable function!

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Stephen Tashi
Let $g(x,y)=\sqrt[3]{xy}$

a) Is $g$ continuous at $(0,0)$?

3. The Attempt at a Solution
a) $$lim_{y \rightarrow 0 \space along\space x=0}\space g(x,y) =0$$

$$lim_{x \rightarrow 0 \space along\space y=0}\space g(x,y)=0$$

$$lim_{y=mx\rightarrow 0} = 0$$

Therefore limit exists and the limit is $0$.
The problem didn't ask you to prove that g(x,y) is continuous at (0,0), but if you are going to present an argument that it is, you should quote the theorems that you use in your argument. You don't want to give the impression that your argument is based on the equality of two interated limits, because that's not a valid argument. (See example 2 in https://en.wikipedia.org/wiki/Iterated_limit ).

It isn't clear what your notation "$lim_{y=mx\rightarrow 0} g(x,y)$ means. You might be trying to convey the idea that, for each number m, $lim_{x \rightarrow 0} g(x,mx) = 0$.

In the first place, what is the proof of that statement? And in the second place, if that statement is true, is there a theorem that concludes $\lim_{(x,y)\rightarrow (0,0)} g(x,y) = 0$ ?

If your text hasn't proven any useful theorems and you wish to prove $\lim_{(x,y)\rightarrow (0,0)} g(x,y) = 0$ then you'll have to begin with the definition of that sort of limit - which (as the link above points out) is not the same as the definition of the two iterated limits.

You cannot assume that $\lim_{x,y \to 0} g_x(x,y)$ is $g_x(0,0)$. The actual definition of $g_x(0,0)$ is
$$g_x(0,0) = \lim_{h \to 0} \frac{g(h,0) - g(0,0)}{h}$$
if the limit exists.

For multivariate functions, what is the actual definition of the term "differentiable"? Be warned: it may not be the same as for a single-variable function!
Ok, let me give this question another shot. I've been drowning under concepts such as continuity and differentiability. So, I've done some reading and will try again.

a) $g$ is continuous at $(0,0)$ because the function is defined at $(0,0)$, and as above, the limit exists and is equal to zero. (Please correct me if I'm wrong)

c) May I know why I cannot assume that $\lim_{x,y\to 0}g_{x}(0,0)$ is $g_x(0,0)$?
So, following what you wrote, $$g_x(0,0)=\lim_{h \to 0}\frac{g(h,0) - g(0,0)}{h}$$
$$=\frac{0-0}{h}$$
which is equal to zero.
The same result is reached for $g_y(0,0)$

d) Since the limits of $\frac {\partial g}{\partial x}$ and $\frac {\partial g}{\partial y}$ are zero and equal to $h \to 0$, I say that both partials are continuous at $(0,0)$.

e) I think this concept involves differentiability, and the question I should ask is does $$lim_{(x,y) \to (0,0)} \frac{g(x,y)- 0}{\|(x,y)-(0,0)\|}=0$$

Along the line $y=0$, $$\frac{g(x,y)}{\|(x,y)\|}=\frac{0}{\sqrt x^2}=0$$

Along the line $y=mx\to 0$, $$\frac{g(x,mx)}{\|(x,mx)\|}=\frac{(mx^2)^\frac{1}{3}}{\sqrt(x^2+m^2x^2)}$$

$$=\frac{m^\frac{1}{3}x^\frac{2}{3}}{x\sqrt(1+m^2)}$$

which is undefined.

Therefore, there is no tangent plane at $(0,0)$

e) Since there is no tangent plane at $(0,0)$, $g$ is not differentiable at that point.

I really hope I'm right this time.

I've read up on continuity and differentiability, and for multivariate functions, the function is differentiable if first there exist a tangent plane equation and more importantly that $$lim_{(x,y) \to (a,b)} \frac {f(x,y)-h(x,y)}{\|(x,y)-(a,b)\|}=0$$.

But I don't understand why.

Ray Vickson
Homework Helper
Dearly Missed
Ok, let me give this question another shot. I've been drowning under concepts such as continuity and differentiability. So, I've done some reading and will try again.

a) $g$ is continuous at $(0,0)$ because the function is defined at $(0,0)$, and as above, the limit exists and is equal to zero. (Please correct me if I'm wrong)

c) May I know why I cannot assume that $\lim_{x,y\to 0}g_{x}(0,0)$ is $g_x(0,0)$?
So, following what you wrote, $$g_x(0,0)=\lim_{h \to 0}\frac{g(h,0) - g(0,0)}{h}$$
$$=\frac{0-0}{h}$$
which is equal to zero.
The same result is reached for $g_y(0,0)$

d) Since the limits of $\frac {\partial g}{\partial x}$ and $\frac {\partial g}{\partial y}$ are zero and equal to $h \to 0$, I say that both partials are continuous at $(0,0)$.
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No, absolutely not!
$g_x(x,y) = (1/3) y^{1/3}/x^{2/3}, \; g_y(x,y) = (1/3) x^{1/3}/y^{2/3}$ for $x,y > 0$. The value of $\lim_{x,y \to 0} g_x(x,y)$ depends very much on how $x$ and $y$ approach 0. If we take $y = k x$ for some constant $k > 0$ we have $g_x(x,y) = g_x(x,kx) =(1/3) k^{1/3}/x^{1/3}$. For each $k > 0$ the limit $\lim_{x \to 0} g_x(x,kx)$ does not exist; certainly, it is not equal to zero!

***********************
e) I think this concept involves differentiability, and the question I should ask is does $$lim_{(x,y) \to (0,0)} \frac{g(x,y)- 0}{\|(x,y)-(0,0)\|}=0$$

Along the line $y=0$, $$\frac{g(x,y)}{\|(x,y)\|}=\frac{0}{\sqrt x^2}=0$$

Along the line $y=mx\to 0$, $$\frac{g(x,mx)}{\|(x,mx)\|}=\frac{(mx^2)^\frac{1}{3}}{\sqrt(x^2+m^2x^2)}$$

$$=\frac{m^\frac{1}{3}x^\frac{2}{3}}{x\sqrt(1+m^2)}$$

which is undefined.

Therefore, there is no tangent plane at $(0,0)$

e) Since there is no tangent plane at $(0,0)$, $g$ is not differentiable at that point.

I really hope I'm right this time.

I've read up on continuity and differentiability, and for multivariate functions, the function is differentiable if first there exist a tangent plane equation and more importantly that $$lim_{(x,y) \to (a,b)} \frac {f(x,y)-h(x,y)}{\|(x,y)-(a,b)\|}=0$$.
**************************
This should be
$$\lim_{(x,y) \to (a,b)} \frac {f(x,y)-f(a,b)-A(x-a)-B(y-b)}{\|(x,y)-(a,b)\|}=0$$
for some finite real constants $A,B$ (which might be functions of the point $(a,b)$). In plain English, this says that for small $|x-a|$ and $|y-b|$ we have
$$f(x,y) = f(a,b) + A(x-a) + B(y-b) + \text{small error},$$
where the "small error" goes to zero faster than linear in $(x-a)$ and $(y-b)$; that is,
$$\frac{\text{small error}}{\|(x,y) - (a,b)\| } \to 0$$
as $(x,y) \to (0,0)$. Of course, we have $A = f_x(a,b)$ and $B = f_y(a,b)$.

Geometrically, this says that the difference between the surface $z = f(x,y)$ and its tangent plane $z = f(a,b) + A(x-a) + y(x-b)$ is of "higher order in small quantities". The exact surface is very nearly flat when we stay near enough to $(a,b)$.

*****************************

But I don't understand why.
Reason? Because it is definition. End of story.

SammyS
Staff Emeritus
Homework Helper
Gold Member
From OP, your solution to part (a): (With some typos fixed)
3. The Attempt at a Solution
a)
$\displaystyle \lim_{y \rightarrow 0 \space \text{along}\space x=0}\space g(x,y) =0$

$\displaystyle \lim_{x \rightarrow 0 \space \text{along}\space y=0}\space g(x,y)=0$

$\displaystyle \lim_{y=mx\rightarrow 0} g(x,y)= 0$​

Therefore limit exists and the limit is 0.
You should know from some of your recent threads, that this is not sufficient to show the limit exists and is zero. (Of course, the limit is 0, but this does not prove it.)
See @Ray Vickson 's post. (Link) for a good method for this function.

***********************
No, absolutely not!
$g_x(x,y) = (1/3) y^{1/3}/x^{2/3}, \; g_y(x,y) = (1/3) x^{1/3}/y^{2/3}$ for $x,y > 0$. The value of $\lim_{x,y \to 0} g_x(x,y)$ depends very much on how $x$ and $y$ approach 0. If we take $y = k x$ for some constant $k > 0$ we have $g_x(x,y) = g_x(x,kx) =(1/3) k^{1/3}/x^{1/3}$. For each $k > 0$ the limit $\lim_{x \to 0} g_x(x,kx)$ does not exist; certainly, it is not equal to zero!

***********************
This question really gives me a headache. So how do I go about showing that $g_x(0,0)$ and $g_y(0,0)$ exist by supplying values for them? Is it because the limit doesn't actually exist that I must 'supply' values for them? I'm so confused now.

Mark44
Mentor
This question really gives me a headache. So how do I go about showing that $g_x(0,0)$ and $g_y(0,0)$ exist by supplying values for them? Is it because the limit doesn't actually exist that I must 'supply' values for them? I'm so confused now.
In post 1 you have formulas for the two first partials. Clearly they aren't defined at (0, 0).
toforfiltum said:
b) $$\frac {\partial g}{\partial x}=\frac{\frac{1}{3}y^\frac{1}{3}}{x^\frac{2}{3}}$$
$$\frac{\partial g}{\partial y}=\frac{\frac{1}{3}x^\frac{1}{3}}{y^\frac{2}{3}}$$

In post 1 you have formulas for the two first partials. Clearly they aren't defined at (0, 0).
Ok, thanks. Now I have a question. When the partials aren't defined at a certain point, must we use the definition as given in post #2 to supply values for them?

Ray Vickson
Homework Helper
Dearly Missed
This question really gives me a headache. So how do I go about showing that $g_x(0,0)$ and $g_y(0,0)$ exist by supplying values for them? Is it because the limit doesn't actually exist that I must 'supply' values for them? I'm so confused now.
You see if
$$\lim_{h \to 0} \frac{g(h,0) - g(0,0)}{h}$$
exists, or not. If it exists, it is by definition, $g_x(0,0)$.

Note: that is different from the question of whether or not $g_x(x,y)$ has a limit as $(x,y) \to (0,0)$ and if so, whether or not that limit equals $g_x(0,0)$.

Stephen Tashi
a) $g$ is continuous at $(0,0)$ because the function is defined at $(0,0)$, and as above, the limit exists and is equal to zero. (Please correct me if I'm wrong)
You are correct that $g$ is continuous at $(0,0)$ and that $g(0,0) = 0$. It isn't clear what limit you mean by "the limit". If you mean $\lim_{(x,y)\rightarrow(0,0)} g(x,y)$ then that is correct limit. You haven't given any proof that the value of that limit is zero, but I don't know whether question a) expected you to give any proof.

c) May I know why I cannot assume that $\lim_{x,y\to 0}g_{x}(0,0)$ is $g_x(0,0)$?
The limit of $\lim_{(x,y)\rightarrow 0}g_x(0,0)$ denotes the limit of a constant function. If we can show that $g_x(0,0) = 0$ then that limit is literally $\lim_{(x,y)\rightarrow 0}0 = 0$. However question c) asked that you show $g_x(0,0) = 0$ You can't show that by beginning with the assumption that $g_x(0,0) = 0$.

So, following what you wrote, $$g_x(0,0)=\lim_{h \to 0}\frac{g(h,0) - g(0,0)}{h}$$
$$=\frac{0-0}{h}$$
which is equal to zero.
Ok, but that would be more explanatory if you wrote $g()$ as a specific function.

d) Since the limits of $\frac {\partial g}{\partial x}$ and $\frac {\partial g}{\partial y}$ are zero and equal to $h \to 0$, I say that both partials are continuous at $(0,0)$.
What limits do you mean by "the limits"? and what do you mean by "equal to $h \to 0$"?

The relevant limit for the continuity of $g_x$ would be $\lim_{(x,y)\rightarrow (0,0)} g_x(x,y)$, which could be made more specific by using the answer to part b) in place of $g_x(x,y)$. I think you gave a correct answer to part d) in your original post.

e) I think this concept involves differentiability, and the question I should ask is does $$lim_{(x,y) \to (0,0)} \frac{g(x,y)- 0}{\|(x,y)-(0,0)\|}=0$$
It's very tempting to formulate the two dimensional definition for a derivative that way. But think about the generalization of a derivative (slope of a line) to the case of the "slope of a plane". The problem is that a plane doesn't have a single slope. If you were walking over a point on a sloped planar parking lot, you would encounter one slope walking "uphill" and a zero slope walking sideways to that direction. Trying to define the derivative as $lim_{(x,y) \to (0,0)} \frac{g(x,y)- 0}{\|(x,y)-(0,0)\|}$ defines something that only exists if the value of that limit didn't depend on the direction that (x,y) approaches (0,0). To capture the idea of a tangent plane, we do want a concept where the slope depends on direction.

The definition of differentiability for a function of two variables is based on the idea that a change in the function from (x,y) to (x+dx y+dy) is approximated by a linear function,and the value of the linear function depends on the direction (dx,dy). The value of the linear function is typically not constant with respect to the length of the vector (dx,dy).

Therefore, there is no tangent plane at $(0,0)$
Investigating the tangent plane is the correct approach, but you weren't looking at the correct limit.

@Stephen Tashi @Ray Vickson Thanks for all your time and explanations! Will go over what you have wrote after my brain is fresh again.

a) Following the recommendation to convert to polar coordinates, $$g(r \cos\theta, r\sin\theta)=(r^2 \cos\theta\sin\theta)^\frac{1}{3}$$
$$\lim_{r \to 0}r^\frac{2}{3}\sqrt[3]{\cos\theta\sin\theta}=0$$
Since $g(0,0)=0$ and $\lim_{r \to 0}g(r\cos\theta\sin\theta)=0$, $g$ is continuous at $(0,0)$.

b) It is correct in the OP, but I'll just put it here again
$$\frac {\partial g}{\partial x}=\frac{\frac{1}{3}y^\frac{1}{3}}{x^\frac{2}{3}}$$
$$\frac{\partial g}{\partial y}=\frac{\frac{1}{3}x^\frac{1}{3}}{y^\frac{2}{3}}$$

c) By definition, $$g_x(0,0)=\lim_{h \to 0}\frac{g(h,0)-g(0,0)}{h}$$
$$=0$$
$$g_y(0,0)=\lim_{h \to 0}\frac{g(0,h)-g(0,0)}{h}$$
$$=0$$

d) As @Ray Vickson said in #5, limit does not exist if we approach it along the line $y=kx$. I would just like to clear up some things here. If I approach it along the line $y=0$, as $x \to 0$, the limit is also undefined, right? Same thing happens if I approach it along the line $x=0$. Since, the limit is undefined for $\frac{\partial g}{\partial x}$ at $(0,0)$ and not equal to zero, I can say that it is not continuous at $(0,0)$. The same for $\frac{\partial g}{\partial y}$.

e) @Stephen Tashi I don't really understand what you wrote in the post above. You mean the direction has something to do with the normal vector of the tangent plane? But since, the partials are not continuous and not defined at $(0,0)$, there is no tangent plane at that point. I think you meant this limit, right?

f) Since $g$ is not continuous at $(0,0)$, therefore is is not differentiable at that point.

Just one last observation, although the partials may not be defined at a certain point, by the definition given in #5, the partial could exist by that definition, is it?

Thanks all.

Stephen Tashi
a) Following the recommendation to convert to polar coordinates,
The context of that recommendation was in showing a function was not continuous.

$$g(r \cos\theta, r\sin\theta)=(r^2 \cos\theta\sin\theta)^\frac{1}{3}$$
$$\lim_{r \to 0}r^\frac{2}{3}\sqrt[3]{\cos\theta\sin\theta}=0$$
Since $g(0,0)=0$ and $\lim_{r \to 0}g(r\cos\theta\sin\theta)=0$, $g$ is continuous at $(0,0)$.
I'd like to hear the comments of the other advisors on whether there is a theorem that allows us to conclude that $g(x,y)$ is continuous from the above calculations.

The above calculations at least show that $lim_{x \rightarrow 0} g(x,kx) = 0$ for any path $y = kx$. However, there are examples where that result is not sufficient to prove continuity. This page http://www.mathcounterexamples.net/continuity-multivariable-real-functions/ has an example of a function $g(x,y)$ where the limit along any line through the origin exists and is zero $= g(0,0)$, but the function is discontinuous at the origin because along the curved line $(x,x^2)$ the limit of $g$ at the origin is 1/2.

So the question is whether the polar coordinate calculations imply more that a statement about limits along straight lines.

d) As @Ray Vickson said in #5, limit does not exist if we approach it along the line $y=kx$.
As a matter of style in writing about mathematics, you shouldn't use vague references like "the limit" and "approach it" because it isn't clear what limit you are talking about and what "it" refers to.

e) @Stephen Tashi I don't really understand what you wrote in the post above.
My remarks have to do with the fact that you used the wrong definition for the derivative of a function of two variables. Your definition for the derivative of $g(x,y)$ at $(0,0)$ was $\lim_{(x,y)\rightarrow (0,0)} \frac{g(x,y) - 0}{||(x,y) - (0,0)||}$.

You have the intuitive idea of the derivative of function of two variables as having something to do with a tangent plane, but you need to understand the formal definition. (Ray Vickson stated it.)

You should refer to specific theorems in your text material to justify your work. As a simple example, what theorem are you going to use to show that $g(x,y)$ is not differentiable at $(0,0)$? You can make intuitive arguments about a tangent plane not existing, but how would you prove it to someone who demanded to see the justification as a theorem in your textbook ?

f) Since $g$ is not continuous at $(0,0)$, therefore is is not differentiable at that point.
No, you argued in question a) that $g$ is continuous at $(0,0)$

Just one last observation, although the partials may not be defined at a certain point, by the definition given in #5, the partial could exist by that definition, is it?
I think what you mean is that the partial derivatives $g_x, g_y$ in this problem are given by formulas that apply when $xy \ne 0$ and these formulas do not have a defined value at $(0,0)$. However, the definition of partial derivative allows us to compute the values of $g_x(0,0), g_y(0,0)$ by using the limits that define the value of these partial derivatives at $(0,0)$. That is correct.

Ray Vickson
Homework Helper
Dearly Missed
@Stephen Tashi @Ray Vickson Thanks for all your time and explanations! Will go over what you have wrote after my brain is fresh again.
If you think that the definition I gave (standard, not mine!) in post #5 is mysterious and elaborate, you should realize that, geometrically speaking, it is very similar to the concept of differentiability of a single-variable function! In one variable, we say that $f(x)$ is differentiable at $x = x_0$ if, near $x = x_0$, the graph of $y = f(x)$ is almost a straight line (i.e, is flat). In other words, there is a constant $A$ such that $f(x_0+h) = f(x_0) + h A + o(|h|)$. Here the notation $o(|h|)$ is shorthand for "terms of higher order in small $|h|$"; that is, $\lim_{h \to 0} o(|h|)/h = 0$. The notion of differentiability in two variables is essentially the same: the graph of the surface $z = f(x,y)$ near $(x,y) = (x_0,y_0)$ is almost flat.

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No, you argued in question a) that g is continuous at (0,0)
Oops, really sorry for my error there. It is not differentiable at that point because the partial derivatives do not exist at $(0,0)$.

how would you prove it to someone who demanded to see the justification as a theorem in your textbook ?
There is a definition in my textbook that states that a function $f$ is differentiable at $(a,b)$ if the partial derivatives $f_x$ and $f_y$ exist and if the function $$h(x,y)=f(a,b) + f_x(a,b)(x-a) +f_y(a,b)(y-b)$$ is a good linear approximation to $f$ near $(a,b)$, that is, if $$\lim_{(x,y)\to (a,b)} \frac{f(x,y)-h(x,y)}{\|(x,y)-(a,b)\|}=0$$

I copied this directly from my textbook.
Is this good enough?

Stephen Tashi
Oops, really sorry for my error there. It is not differentiable at that point because the partial derivatives do not exist at $(0,0)$.
But you showed that $g_x(x,y)$ and $g_y(x,y)$ do exist at $(0,0)$ , as instructed in part c).

You also showed $g_x(x,y)$ and $g_y(x,y)$ are not continuous at $(0,0)$ but don't let that mislead you into thinking that this result proves $g(x,y)$ is not differentiable at $(0,0)$. (See http://mathinsight.org/differentiable_function_discontinuous_partial_derivatives )

There is a definition in my textbook that states that a function $f$ is differentiable at $(a,b)$ if the partial derivatives $f_x$ and $f_y$ exist and if the function $$h(x,y)=f(a,b) + f_x(a,b)(x-a) +f_y(a,b)(y-b)$$ is a good linear approximation to $f$ near $(a,b)$, that is, if $$\lim_{(x,y)\to (a,b)} \frac{f(x,y)-h(x,y)}{\|(x,y)-(a,b)\|}=0$$
That definition looks ok to me. If you have to prove rigorously that $g(x,y)$ is not differentiable at $(0,0)$ you can try using that definition.

From the wording of part e), your instructor may not require a rigorous proof since part e) suggests drawing a graph. Do you know whether the instructor expects a proof or whether an intuitive answer is good enough?

Mark44
Mentor
The thing that bothers me about this problem is the need to use the definition of the partial derivative instead of the rules that are derived from this definition. Using the usual rules, I get $g_x(x, y) = \frac {y^{1/3}}{3x^{2/3}}$. From this I would conclude that $g_x$ is not defined at (0, 0). OTOH, by using the definition of the partial wrt x, we conclude that $g_x(0, 0)$ is defined, and equals 0. Why the disparity? Is there some subtlety I'm missing here?

The thing that bothers me about this problem is the need to use the definition of the partial derivative instead of the rules that are derived from this definition. Using the usual rules, I get $g_x(x, y) = \frac {y^{1/3}}{3x^{2/3}}$. From this I would conclude that $g_x$ is not defined at (0, 0). OTOH, by using the definition of the partial wrt x, we conclude that $g_x(0, 0)$ is defined, and equals 0. Why the disparity? Is there some subtlety I'm missing here?
I just got back from my professor's office hours, and I happened to ask the same question as you are asking now. He mentioned something along the lines of:

Calculating a partial derivative involves two steps, and in a 'nice' function, you could take any step first and it will be fine. But it is not the case for this function.

About the two steps. Firstly, it has been a lot of questions, so I can't quite remember, but he said something like if following the definition, we calculate the partials by first restricting it along a certain plane, then plug in the values.

If we just plug in the values for the partials derived in (b), then we are doing the steps backwards, which just doesn't work for this function, because it is not 'nice'.

Now, I know that what I'm saying here is not clear at all, but I'm hoping that some other members here can 'get' what my professor is talking about, because honestly, I don't understand what he is saying.

SammyS
Staff Emeritus
Homework Helper
Gold Member
@Stephen Tashi ,
a) Following the recommendation to convert to polar coordinates,
The context of that recommendation was in showing a function was not continuous.
$\displaystyle g(r \cos\theta, r\sin\theta)=(r^2 \cos\theta\sin\theta)^\frac{1}{3}$​

$\displaystyle \lim_{r \to 0}r^\frac{2}{3}\sqrt[3]{\cos\theta\sin\theta}=0$​

Since $\ g(0,0)=0\$ and $\ \displaystyle \lim_{r \to 0}g(r\cos\theta,\,r\sin\theta)=0\,,\ g\$ is continuous at (0,0)(0,0).
I'd like to hear the comments of the other advisors on whether there is a theorem that allows us to conclude that $g(x,y)$ is continuous from the above calculations.

The above calculations at least show that $lim_{x \rightarrow 0} g(x,kx) = 0$ for any path $y = kx$. However, there are examples where that result is not sufficient to prove continuity. This page http://www.mathcounterexamples.net/continuity-multivariable-real-functions/ has an example of a function $g(x,y)$ where the limit along any line through the origin exists and is zero $= g(0,0)$, but the function is discontinuous at the origin because along the curved line $(x,x^2)$ the limit of $g$ at the origin is 1/2.

So the question is whether the polar coordinate calculations imply more that a statement about limits along straight lines.

...
If you're referring to my post, the context was to show that g is continuous at (0, 0).

Since sin(θ)⋅cos(θ) is bounded, it follows that $\ \displaystyle \lim_{r \to 0}r^\frac{2}{3}\sqrt[3]{\cos\theta\sin\theta}=0 \ .$ (Use squeeze theorem.) Polar coordinates do work well to show the this limit is zero.

In the case you bring up in that link, it is true that polar coordinates are not very helpful. That function is given by:
$\begin{array}{l|rcl} f : & \mathbb R^2 & \longrightarrow & \mathbb R \\ & (x,y) & \longmapsto & \frac{x^2 y}{x^4+y^2} \text{ for } (x,y) \neq (0,0)\\ & (0,0) & \longmapsto & 0 \end{array}$
(I used f rather than g to avoid confusion.)​
In general, the limit as (x,y) → (0,0) is undefined.
However, if the approach to (0,0) is made along any (straight) line, that limit exists and is zero.

Using polar coordinates at other than the origin:
$\displaystyle f(r\cos(\theta),\, r\sin(\theta))=\frac{r\cos^2(\theta)\sin(\theta)}{r^2\cos^4(\theta)+\sin^2(\theta)}$​
This isn't all that helpful in determining the limit approaching the origin, i.e. r → 0 .

It looks perhaps that the limit is zero. However, try approaching the origin along the circle, r = sin(θ) .

But you showed that $g_x(x,y)$ and $g_y(x,y)$ do exist at $(0,0)$ , as instructed in part c).

You also showed $g_x(x,y)$ and $g_y(x,y)$ are not continuous at $(0,0)$ but don't let that mislead you into thinking that this result proves $g(x,y)$ is not differentiable at $(0,0)$. (See http://mathinsight.org/differentiable_function_discontinuous_partial_derivatives )

That definition looks ok to me. If you have to prove rigorously that $g(x,y)$ is not differentiable at $(0,0)$ you can try using that definition.

From the wording of part e), your instructor may not require a rigorous proof since part e) suggests drawing a graph. Do you know whether the instructor expects a proof or whether an intuitive answer is good enough?
He does not expect a proof. I went to his office hours today, and he asked me to provide an answer along the lines of weird behaviour, involving verticality when approaching $(0,0)$.

Anyway, I plotted the graph using wolfram alpha, and it is evident that a tangent plane does not exist at $(0,0)$. Maybe that's all that is needed. Since no tangent plane exists at that point, $g$ is not differentiable at $(0,0)$.

By the way, thanks for the link you gave me! I've proceeded to read a few other pages regarding differentiability and the differentiability theorem. This website has helped me loads, though I'm afraid not much for this particular question.

Thanks again!

Ray Vickson
Homework Helper
Dearly Missed
The thing that bothers me about this problem is the need to use the definition of the partial derivative instead of the rules that are derived from this definition. Using the usual rules, I get $g_x(x, y) = \frac {y^{1/3}}{3x^{2/3}}$. From this I would conclude that $g_x$ is not defined at (0, 0). OTOH, by using the definition of the partial wrt x, we conclude that $g_x(0, 0)$ is defined, and equals 0. Why the disparity? Is there some subtlety I'm missing here?
The problem is that the notion of differentiability is more elaborate for multivariate functions than for univariate functions. As I hinted in post #15 (explaining post #5 more intuitively), differentiabilty is essentially "local flatness"---in both 1 or several dimensions. The difference is that in 1 dimension, flatness = essentially linear, while in several dimensions, flatness = essentially planar (or hyperplanar, if there is such a term).

Existence of partial derivatives is different: it requires local flatness in two directions only (East-West and North-South); differentiability requires local flatness in all directions, and more. There are examples of functions that are non-continuous at a point $(x_0,y_0)$ while having well-defined, finite partial derivatives $f_x(x_0,y_0)$ and $f_y(x_0,y_0)$, while differentiable functions are provably continuous at $(x_0,y_0)$. There is a theorem that says $f$ is differentiable at $(x_0,y_0)$ if it is continuous at $(x_0,y_0)$ and has partial derivatives $f_x(x,y), f_y(x,y)$ that are continuous in a neighborhood of $(x_0,y_0)$. (I don't recall whether the requirement of continuity of $f$ at $(x_0,y_0)$ can be dropped, or whether it is truly needed. Maybe continuity of the partials implies it already; I just don't have the statement readily at hand.)

Stephen Tashi
Using polar coordinates at other than the origin:
$\displaystyle f(r\cos(\theta),\, r\sin(\theta))=\frac{r\cos^2(\theta)\sin(\theta)}{r^2\cos^4(\theta)+\sin^2(\theta)}$​
This isn't all that helpful in determining the limit approaching the origin, i.e. r → 0 .
That brings up the technicality of how a person would interpret the notation $\lim_{r\rightarrow 0}\frac{r\cos^2(\theta)\sin(\theta)}{r^2\cos^4(\theta)+\sin^2(\theta)}$. A common interpretation would be to treat $\theta$ as a constant. Then by considering two cases i) $\sin{\theta} = 0$ and $\sin{\theta} \ne 0$ the limit would be investigated.

It looks perhaps that the limit is zero. However, try approaching the origin along the circle, r = sin(θ) .
How do we denote a limit where that possibility must be considered? It isn't necessarily the joint limit $\lim_{(r,\theta) \rightarrow (0,0)}\frac{r\cos^2(\theta)\sin(\theta)}{r^2\cos^4(\theta)+\sin^2(\theta)}$ because there is no requirement that $\theta \rightarrow 0$.

Returning to proof:
Since sin(θ)⋅cos(θ) is bounded, it follows that $\ \displaystyle \lim_{r \to 0}r^\frac{2}{3}\sqrt[3]{\cos\theta\sin\theta}=0 \ .$ (Use squeeze theorem.) Polar coordinates do work well to show the this limit is zero.
I agree that polar coordinates can be used, but it seems to me that establishing that $\lim_{r \to 0}r^\frac{2}{3}\sqrt[3]{\cos\theta\sin\theta}=0$ is only one step on the way. I think the proof needs to be finished up by using the definition of $\lim_{(x,y)\rightarrow (0,0)} g(x,y)= 0$.

- either that or define some new type of limit that will exist iff $\lim_{(r,p(r))\rightarrow (0,p(0))} g(r,\theta) = L$ for each possible function $p(r)$ and create a theorem based on that kind of limit instead of the $r\rightarrow 0$ kind.

Stephen Tashi
If we want to check this out, we can try to develop the formula for $\frac{\partial}{\partial x} (xy)^{\frac{1}{3}}$ starting with the definition of derivative and see if there is some place in our work where we must say "if $xy\ne 0$...". If we find that place then to do a complete job we should consider two cases i) $xy \ne 0$ and ii) $xy = 0$. I think we memorize the formula for taking partial derivatives in case i) but forget case ii).