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Given that $f(0,0) = 0$:

How is it possible that $$\frac{∂f}{∂x}(0,0)=\lim\limits_{h\to 0}\frac1h\big(f(0+ h,0)−f(0,0)\big)=\lim\limits_{h\to0}\frac1h\!\cdot\!0=0$$

I assume they did $$\frac{\partial f}{\partial x}(0,0)=\lim\limits_{h\to0}\frac1h\big(f(0+h,0)−f(0,0)\big)=$$ $$=\lim\limits_{h\to 0}\frac1h\cdot\lim\limits_{h\to0}\big(f(0+h,0)−f(0,0)\big)=$$ $$=\lim\limits_{h\to0}\frac1h\!\cdot\!0=0$$

But $$\lim\limits_{h\to0}\frac1h=\infty$$

So how does zero times infinity equal zero? I thought multiplication with infinity was undefined.

Edit: The suggested post doesn't answer my question since my assumption is wrong. I wanted help with understanding how the partial derivative with respect to x is 0. Since this site requires us to show our own work I had to show you my take on it which is clearly wrong.The title has also been edited now to better suit the actual problem.

Need_MathHelp
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    Isn't it $\frac{f(h,0)}{h}$ instead of $\frac1h$ ? Then the limits depends on $f$. – Lelouch May 26 '23 at 13:52
  • @Lelouch what do you mean? – Need_MathHelp May 26 '23 at 14:00
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    You can't say anything about the limit of $f$ at $(0,0)$ if all you know is the value of $f$ at $(0,0)$. There must be information that you have not told us. – David K May 26 '23 at 14:03
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    Just to be clear, $\frac{\partial f}{\partial x} (0, 0) = \lim_{h \to 0}\left( \frac1{h} (f(0 + h, 0) − f(0, 0)) \right)$. It's not $\left(\lim_{h \to 0} \frac1{h}\right) (f(0 + h, 0) − f(0, 0))$. – David K May 26 '23 at 14:08
  • @DavidK Aha no in an earlier question, I have proven that f is continuous at (0,0) – Need_MathHelp May 26 '23 at 14:18
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    @Need_MathHelp: That's irrelevant (or at least not enough information). The missing information that David K is talking about must be that $f(h,0)=0$ for all $h$ (not just for $h=0$). Otherwise your first calculation doesn't make sense. (And your second calculation never makes sense, since the “product of the limits” rule doesn't apply to the case “$\infty \cdot 0$”.) – Hans Lundmark May 26 '23 at 14:57
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    And by the way, $\lim_{h \to 0} (1/h)$ is not equal to $\infty$ (in the context of the extended real line). It doesn't exist, since $\lim_{h \to 0^-} (1/h) = -\infty$ and $\lim_{h \to 0^+} (1/h) = +\infty$ are unequal. – Hans Lundmark May 26 '23 at 14:59
  • "... in an earlier question, I have proven ..." And you didn't think we deserved to see a link to that question? Did you expect us to guess that such a question existed and go hunt it down, or perhaps use telepathy? Next time, link to the earlier question within the new question, like this: How to check if the limit exists in multivariable calculus. – David K May 26 '23 at 16:02
  • In the linked question, $f(0+h,0)=0$ also for all $h$, which is the missing information. – David K May 26 '23 at 16:07
  • @DavidK No I was confused, looking back at the question I totally did something else (never proved that $f(0+h, 0)$ goes to zero'9 so now I am even more confused as to how they got $\frac{\partial f}{\partial x} = 0$. For the record my earlier post "How to check if the limit exists in multivariable calculus" is another question not related to this one. – Need_MathHelp May 27 '23 at 20:57
  • I wrote, "In the linked question, $f(0+h,0)=0$ also for all $h$," not because it the other question literally said so but because I looked at the function defined in that question and was able to conclude that fact. But I did jump to the conclusion that the question you posted before was about the same function $f$ and was therefore related to this one. Is this question about a new function? If so, we really need to know what that function is. I see you've been active here and you seem to know how to ask a good question; I'm surprised that you're having such difficulty asking this one. – David K May 28 '23 at 14:13
  • You are right, I usually post the entire question so it doesn't become the famous "xy-problem". But this time I was only confused about this part of the problem and forgot to realize that you guys need more than just this to work with. Sorry for the confusion Ig:) I've been solving a lot of problems lately so I don't even know where I found this question and thus I can't really edit it. But I stumbled on a similar question where this was not the case so I just wanted once again to confirm if my assumption of what they did was wrong? – Need_MathHelp May 28 '23 at 15:11
  • It's unfortunate that you've lost track of the source of this particular question. But if the source is lost, it is lost, and there is not much you can do about that. The only thing I can think that is left to say is that it is unlikely that someone teaching math would make the mistake of factoring a limit into the product of an undefined limit and zero. It is much more likely that $f(0+h,0)=0$ for $h$ in some neighborhood of $0$ (and only slightly less likely that the neighborhood is all of $\mathbb R$), which leads to $\lim_{h\to0}\left(\frac1h\cdot0\right)=\lim_{h\to0}0=0.$ – David K May 29 '23 at 05:09

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