5
$\begingroup$

The function in question is

$$f(x,y) = \begin{cases} 0 & (x,y)=(0,0)\\ \frac{xy}{|x|+|y|} & (x,y) \neq (0,0) \end{cases} $$

So far I've calculated the continuity of the function at $(0,0)$ using the squeeze theorem with the functions $xy$ and $\frac{xy}{\sqrt{x^2+y^2}}$. The former is always bigger than the second branch of $f$, and the latter always smaller; they all have limit $0$, therefore the function is continuous at $(0,0)$.

Now I have to calculate the first partial derivative in order to $x$ of $f$ at $(0,0)$. Using the definition, we have

$$\frac{\partial f}{\partial x}(0,0)=\lim_{t\to 0}\frac{f(t,0)-f(0,0)}{t}$$

Now, $f(0,0)$ is just $0$, and $f(t,0)$ is $0$ as well. So we have

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

which is not really meaningful, at least to me, and there doesn't seem to be an obvious way of solving this limit without finding the full expression.

So I started hunting for an expression for the partial derivative using the limit definition. That should be

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

but that seems impressingly difficult to calculate. If any of this is correct, we have

$$\lim_{t\to 0}\frac{\frac{(x+t)y}{|x+t|+|y|}-\frac{xy}{|x|+|y|}}{t}=\lim_{t\to 0}\frac{(x+t)y}{t(|x+t|+|y|)}-\frac{xy}{t(|x|+|y|)}$$

and after a few "simplifications" (which don't simplify anything at all), we get

$$\lim_{t\to 0} \frac{xy|x|+ty|x|+xy|y|+ty|y|-xy|x+t|-xy|y|}{t(|x+t|+|y|)(|x|+|y|)}$$

and there's where it becomes a huge mess. I think separating $|x+t|$ would help a lot, but I have no idea on how to justify doing that. This analysis class is killing me, even if you can't help me with the problem just recommending a good textbook would help a ton.

Here's the function plotted on WA: https://www.wolframalpha.com/input/?i=f(x,y)+%3D+(xy)%2F(%7Cx%7C%2B%7Cy%7C)

$\endgroup$
7
  • 1
    $\begingroup$ What is your question? $\endgroup$
    – copper.hat
    Jun 12 '18 at 20:28
  • $\begingroup$ What is the partial derivative at this point, and perhaps some clues towards the partial derivative general expression (WolframAlpha says it exists, but I'm not getting it right). I've never seen $\frac0t$ happen in this context. $\endgroup$ Jun 12 '18 at 20:31
  • 3
    $\begingroup$ $\lim_{t\to 0} \frac{0}{t} = \lim_{t\to 0} 0 = 0$, no? $\endgroup$
    – Xander Henderson
    Jun 12 '18 at 20:31
  • $\begingroup$ Note that $f(x,0) = f(0,y) = 0$ for all $x,y$. Hence the partials exist at $(0,0)$. $\endgroup$
    – copper.hat
    Jun 12 '18 at 20:32
  • $\begingroup$ @copper.hat I think you mean $f(x,0) = f(0,y) = 0$. ;) $\endgroup$
    – Xander Henderson
    Jun 12 '18 at 20:32
3
$\begingroup$

Very roughly speaking, the partial derivative $\frac{\partial f}{\partial x}$ represents the change in the value of $f$ in response to a very small change in the value of $x$, assuming that $y$ is fixed. Note that if we fix $y = 0$, then $$ f(x,0) = \frac{x\cdot 0}{|x|+|0|} = 0. $$ Hence if $y=0$, the function is constant with respect to $x$ (and equal to zero). Thus a small change in $x$ will lead to no change in the value of $f$. Therefore it is reasonable to expect that the partial derivative with respect to $x$ will be zero.

More formally, $$ \frac{\partial f}{\partial x}(x,0) = \lim_{t\to 0} \frac{f(t,0) - f(0,0)}{t} = \lim_{t\to 0} \frac{0}{t} = 0, $$ which is what we (heuristically) expect.

$\endgroup$
2
$\begingroup$

I think your confusion is when you said about the limit $$\lim_{t\to0}\frac{f(t,0)-f(0,0)}{t}=\lim_{t\to 0}\frac{0}{t}$$ that it is not really meaningful. Note that in this limit "$t\to0$" means that $t$ approaches zero but is not equal to zero. So for each admissible value of $t$, the expression $\displaystyle\frac{0}{t}$ is perfectly well-defined, and is in fact equal to zero — which makes the limit perfectly meaningful and in fact equal to $0$ as well.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.