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I would like to prove the differentiability of $f(x,y)=|xy|$ at $(0,0)$ using the $\epsilon_1,\epsilon_2$ definition ( from thomas calculus)

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My Working:

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

similarly $f_y (0,0)=0$

$\Delta z= |\Delta x||\Delta y|$

Hence by definition we must represent $\Delta z$ as $\epsilon_1 \Delta x + \epsilon_2 \Delta y $

$|\Delta x||\Delta y|=\epsilon_1 \Delta x + \epsilon_2 \Delta y $

but I am unable to find any such $\epsilon_1,\epsilon_2$ and moreover they must tend to 0 as $\Delta x, \Delta y$ tend to 0. I know you can prove it using the total derivative definition but what's going wrong here?

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    $\begingroup$ Welcome to MSE. From How to ask a good question: Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. $\endgroup$
    – jjagmath
    Dec 31, 2021 at 17:39
  • $\begingroup$ ok i will my edit my question $\endgroup$ Dec 31, 2021 at 17:42
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    $\begingroup$ Find partial derivatives at $(0,0)$. Those are the candidates and prove for $\left(\frac{\partial f}{\partial x}(0,0),\frac{\partial f}{\partial y}(0,0)\right)$. $\endgroup$
    – PinkyWay
    Dec 31, 2021 at 18:25
  • $\begingroup$ Try $\epsilon_1 = |\Delta y| \cdot sign \, \Delta x$ and find an $\epsilon_2$ that works. $\endgroup$ Dec 31, 2021 at 18:30
  • $\begingroup$ $$\lim_{x\to c}\frac{\|f(x)-f(c)-L(x-c)\|}{\|x-c\|}=0$$ is more practical. $\endgroup$
    – PinkyWay
    Dec 31, 2021 at 18:36

2 Answers 2

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I think we can just take $$ \epsilon_1 = \frac{|\Delta x \Delta y|}{2\Delta x}\\ \epsilon_2 = \frac{|\Delta x \Delta y|}{2\Delta y} $$ $\newcommand{\sgn}{\textrm{sgn}}$ Recalling that $|a| = \sgn(a)a$, for all nonzero $\Delta x $ and $ \Delta y$, $$ \epsilon_1 = \frac{1}{2}\sgn(\Delta x)|\Delta y| $$ and $|\epsilon_1| = |\Delta y|/2$. Therefore, $$ \lim_{\Delta x \to 0}\left(\lim_{\Delta y \to 0} \epsilon_1\right) = 0 $$ The process similarly applies to $\epsilon_2$.

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  • $\begingroup$ Even $\epsilon_1 = |\Delta_x \Delta y|/\Delta_x$ and $\epsilon_2 = 0$ would work? $\endgroup$
    – Hermis14
    Dec 31, 2021 at 18:42
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    $\begingroup$ ok got it this works I thought they meant that individually $\lim_{\Delta x\to 0} \epsilon_1$ and $\lim_{\Delta y\to 0} \epsilon_2$. so $\epsilon_1$ and $\epsilon_2$ are functions of both $\Delta x$ and $\Delta y$ right? $\endgroup$ Dec 31, 2021 at 18:47
  • $\begingroup$ @NirmalGovindaraj Yes, they are! $\endgroup$
    – Hermis14
    Dec 31, 2021 at 19:08
  • $\begingroup$ thank you for clarifying this $\endgroup$ Dec 31, 2021 at 19:19
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I would recommend computing partial derivatives and then checking that the limit definition you have works out. In particular, $$f_x(0,0) = \lim_{\Delta x\to 0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x} = \lim_{\Delta x\to 0} \frac 0{\Delta x} = 0,$$ and similarly for $f_y$. So you need to look at $$\Delta z = |\Delta x\Delta y| - 0\Delta x - 0\Delta y = |\Delta x\Delta y|.$$ Now this can be split up in zillions of different ways as $\epsilon_1\Delta x+\epsilon_2\Delta y$. For example, take $\epsilon_1 = \frac12\eta_1|\Delta y|$ and $\epsilon_2 = \frac12\eta_2|\Delta x|$, where $\eta_1 = \dfrac{|\Delta x|}{\Delta x}$ whenever $\Delta x\ne 0$ (and say $0$ when $\Delta x = 0$), and similarly for $\eta_2$. This is actually quite an awkward definition to work with in this case because of the absolute values.

For your reference, a more common definition, which is easier to work with, is that we should have $$\lim_{(\Delta x,\Delta y)\to (0,0)} \frac{\Delta z - f_x(x_0,y_0)\Delta x - f_y(x_0,y_0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0.$$ In our case, we only have to check that $$\lim_{(\Delta x,\Delta y)\to (0,0)} \frac{|\Delta x\Delta y|}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0.$$ Can you do this?

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  • $\begingroup$ yes the second one is much easier using squeeze theorem but I was thinking that $\epsilon_1$ was a function of only $\Delta x$ which is why I couldnt get it but now its clear, Thanks! $\endgroup$ Dec 31, 2021 at 19:23

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