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I'm currently studying Partial Derivatives and ran into this problem:

$f(x,y) = \begin{cases} \frac{x^2}{x^2+y^4}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases}$

In this question, we are supposed to find $\frac{\partial f}{\partial x}$ at $(0,0)$.

(Sorry if this question seems very easy, but I'm new to the material).

From what I know, we use the definition as follows:

$\frac{\partial f}{\partial x}(x_0,y_0) = \lim_{\delta x \to 0} \frac{f(x_0+\delta x,y_0) - f(x_0,y_0)}{\delta x}$.

At $(0,0)$ we have $f(x,y) = 0.$ Therefore,

$\lim_{\delta x \to 0} \frac{0 - 0}{\delta x} = 0$.

Is this correct? Or do we do the following:

$\lim_{\delta x \to 0} \frac{\frac{\delta x^2}{\delta x^2 + 0^4}-\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^4}}{\delta x}$

We can observe that $\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^4}$ does not exist as it converges to different values at different paths (try $x = y$ and $x = y^2$). So, $\frac{\partial f}{\partial x}(0,0)$ does not exist.

Now I believe the first solution is correct because the question says that at $(0,0)$, the function is equal to zero and so we only consider the zero while finding the limit at that point.

But I'm not 100% sure. I talked about this with other students and I found conflicting answers.

Would it be possible for someone to tell me which answer is the correct one? A brief explanation would be very much appreciated.

Thank you.

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  • $\begingroup$ The only paths you would check would be in the positive and negative $x$ directions for the partial in the $x$ variable $\endgroup$
    – Alan
    Aug 2 at 18:06
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By definition,$$\frac\partial{\partial x}f(x_0,y_0)=\lim_{h\to0}\frac{f(x_0+h,y_0)-f(x_0,y_0)}h.$$In particular$$\frac\partial{\partial x}f(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}h.$$But $f(0,0)=0$ and, for each $h\in\Bbb R\setminus\{0\}$, $f(h,0)=\frac{h^2}{h^2}=1$. Therefore, the limit$$\lim_{h\to0}\frac{f(h,0)-f(0,0)}h\quad\text{is}\quad\lim_{h\to0}\frac1h,$$which does not exist (in $\Bbb R$).

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  • $\begingroup$ Sorry if I'm missing something here, but the limit $\lim_{h \to 0}\frac{1}{h}$ doesn't exist at all, right? Even if we adopt a convention where we say that the limit exists when it is $\pm\infty$, it is still not the case that $\lim_{h \to 0}\frac{1}{h}$ exists. $\endgroup$
    – Joe
    Aug 2 at 18:15
  • $\begingroup$ Yes, but we can also add to $\Bbb R$ a single point at infinity (as we do with $\Bbb C$). If we choose to do that, then $\lim_{h\to0}\frac1h=\infty$. $\endgroup$ Aug 2 at 18:17
  • $\begingroup$ Oh, okay, thanks for clarifying. Is the set that you are referring to called the projectively extended real line? $\endgroup$
    – Joe
    Aug 2 at 18:19
  • $\begingroup$ Yes, that is correct. $\endgroup$ Aug 2 at 18:24
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José has given you the correct reasoning using the definition of $\frac{\partial f}{\partial x}$. I would like to constructively critique some of the other methods that you and your classmates used.

First, recall the (somewhat abridged) Algebra of Limits theorem:

Suppose $\lim_{x \to a} f(x) = A$ and $\lim_{x \to a} g(x) = B$. Then

  • $\lim_{x \to a} (f(x) - g(x)) = A - B$
  • $\lim_{x \to a} f(x)/g(x) = A/B$, whenever $B \neq 0$.

So, under the given hypotheses, we can simultaneously substitute each function for its limit. This is the theorem that you're sort of using in your first attempt, but there are two problems here.

The first is that you've substituted $f(x_0 + \delta x, y_0) = f(0 + \delta x, 0)$ for its function value at $\delta x = 0$, not its limit. In the cases where $f$ is continuous at $(0, 0)$, these two quantities will be exactly the same. In this case, $f$ is not continuous at this point. If you substitute anything for that expression, it actually should be $1$, which is the limit of $f(\delta x, 0)$ as $\delta x \to 0$: $$\lim_{\delta x \to 0} f(\delta x, 0) = \lim_{\delta x \to 0} \frac{\delta x^2}{\delta x^2 + 0^4} = \lim_{\delta x \to 0} 1 = 1.$$ (By the by, this shows that $x \mapsto f(x, 0)$ is not continuous at $0$, which means it is not differentiable at $0$. However, this derivative at $0$ is precisely the partial $x$ derivative of $f$, at $(0, 0)$, so this is a quick way of determining the derivative doesn't exist.)

So, let's ignore that first point, and say for the moment that $\lim_{\delta x} f(\delta x, 0) = 0$. Then, it is perfectly valid to say, by the first point of the Algebra of Limits theorem, that $$\lim_{\delta x \to 0} (f(\delta x, 0) - f(0, 0)) = 0 - 0 = 0.$$ So, if we divide $f(\delta x, 0) - f(0, 0)$ by some function $h(\delta x)$ with a non-zero limit $L$, then we can say that $$\lim_{\delta x \to 0}\frac{f(\delta x, 0) - f(0, 0)}{h(\delta x)} = \frac{0}{L} = 0.$$ Our problem is that our actual denominator is $\delta x$, which limits to $0$. So, instead (presumably because it doesn't lead to a division by $0$), you opt for a partial substitution, which is not in the Algebra of Limits theorem! Partial substitutions are sometimes valid to do under certain assumptions (even though this particular theorem doesn't directly sanction them), but as a rule of thumb, if the partial substitution is at all helpful, then it's not valid and probably giving you the wrong answer (as it is in this case)! Basically what I'm saying is, don't make partial substitutions.

Here is a simple example of how it goes wrong. Note that $\frac{x}{x} = 1$ for $x \neq 0$, so $\lim_{x \to 0} \frac{x}{x} = 1$. But, if we opt for a partial substitution, $$\lim_{x \to 0} \frac{x}{x} \stackrel{?}{=} \lim_{x \to 0} \frac{0}{x} = 0.$$ Note: the conclusion is all wrong!


Your second approach is a perfectly valid proof that the limit of $f(x, y)$ does not exist as $(x, y) \to (0, 0)$. If the limit existed, then the limit of the function along different curves through $(0, 0)$ would all be equal to this one limit, but since they come to different quantities, this limit doesn't exist.

Since this limit doesn't exist, $f$ is not continuous at $(0, 0)$, so it doesn't have a total derivative at $(0, 0)$. Unfortunately, this isn't enough to conclude that either of the partial derivatives don't exist! Indeed, even though this function is discontinuous, $\frac{\partial f}{\partial y}(0, 0)$ exists! We have $$\frac{\partial f}{\partial y}(0, 0) = \lim_{\delta y \to 0} \frac{f(0, 0 + \delta y) - f(0,0)}{\delta y} = \lim_{\delta y \to 0} \frac{\frac{0^2}{0^2 + \delta y^4} - 0}{\delta y} = \lim_{\delta y \to 0} \frac{0 - 0}{\delta y} = 0.$$ So, this logic isn't sufficient to say that the $y$ partial derivative doesn't exist, so it's not valid to claim the same for the $x$ partial derivative.

Hopefully some of that helps.

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