Question about existence of partial derivatives at a point

Question

When a function has a problem at a certain point (for example point $(0,0)$ makes the denominator go to zero-I will use this here) and we define it at that point in cases where doing this makes the function continuous, then how to obtain the partial derivatives of the function at that point?

I know that we can do that with the definition of the partial derivatives. But, can't we just find the derivatives of the function for non-problematic points and then take the limit $(x,y)\rightarrow(0,0)$?

I did this is some exercises I found and while I got particular answers using the definition, when I did the second method I found that the limit depended on the parameter m when I put $y=mx$ to find if the limit exists. So, I am mistaken, but I don't know where and why.

EDIT: So, now I know that those two things are not the same. I understand it, formally. But, conceptually I do not understand their difference.

Example(from the comments): For example, the function $f=0$ when $(x,y)=(0,0)$ and $f=(xy^2-x^2y+3x^3-y^3)/(x^2+y^2)$ otherwise. Using the definition of partial derivatives, I find that the two partials are zero at $(0,0)$. But if I compute the functions of the partials and then take their limit as $(x,y)\rightarrow(0,0)$ and put y=mx I find that each limit depends on the value of m; which means that the limits do not exist. And my problem is why are these two ways not equivalent? What's conceptually(or informally) different about them? Why don't I get the same result?

Answer 1

"But, can't we just find the derivatives of the function for non-problematic points and then take the limit (x,y)->(0,0)?". Yes you can. Suppose $f$ is continuous at $(0,0),$ $f_x,f_y$ exist in a deleted neighborhood of $(0,0),$ and

$$\tag 1\lim_{(x,y)\to (0,0)} f_x(x,y) = a, \lim_{(x,y)\to (0,0)} f_y(x,y) = b,$$

where $a,b \in \mathbb R.$ Then $f_x(0,0)=a, f_y(0,0)=b.$ (And in fact $f$ is differentiable at $(0,0)$ with $Df(0,0)(x,y) = ax + by.$) That's a nice little mean value theorem exercise.

But, Warnings: 1) You can't find limits as in $(1)$ by just looking along certain paths. 2) Just because the limits in $(1)$ fail to exist does not imply $f_x(0,0), f_y(0,0)$ fail to exist, nor does it imply $Df(0,0)$ fails to exist.

Answer 2

zhw's answer shows that what you propose is fine provided that the function is sufficiently continuous and well-behaved (note that a function may be wildly discontinuous at $(0,0)$ and still have both partial derivatives exist at $(0,0)$). But this is a theorem, it's something that must be justified (and it can be, given the specified regularity assumptions).

Let's break down what you're saying into formal definitions. The partial derivative $f_x(0,0)$ is defined by the limit:

$$f_x(0,0) = \lim_{y\to 0} \frac{f(0,y) - f(0,0)}{y}.$$

The limit you are asking about in your question is the limit of the partial derivative at values away from $(0,0)$. Specifically:

$$\lim_{(x_0,y_0)\to(0,0)} f_x(x_0,y_0) = \lim_{(x_0,y_0)\to(0,0)} \lim_{y\to y_0} \frac{f(x_0,y) - f(x_0,y_0)}{y-y_0}.$$

There is no intrinsic reason to believe these are the same for arbitrary functions: you can't assume that limits within limits pass through without further conditions on the function $f$.

In the comments I mentioned that the one-variable function $g(x) = x^2 \sin(1/x^2)$ satisfies $g'(0) = 0$ but $\lim_{x\to 0} g'(x)$ does not exist. So the function $f(x,y) = g(x)$ would have the same problem. But there are simpler examples that I alluded to earlier: take any function where $f(x,y) = 0$ whenever $xy=0$, but where $f(x,y)$ takes arbitrary values for $x\ne0$ and $y\ne 0$. Then $f_x(0,0) = f_y(0,0) = 0$, but $f_x(x_0,y_0)$ may very well not exist for any $y_0\ne 0$, let alone the limit of $f_x(x_0,y_0)$.