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Is there a function (non piece-wise unlike below) which is discontinuous but has directional derivative at particular point? I have a manual that says the function has directional derivative at $(0,0)$ but is not continuous at $(0,0)$. $$f(x,y) = \begin{cases} \frac{xy^2}{x^2+y^4} & \text{ if } x \neq 0\\ 0 & \text{ if } x= 0 \end{cases}$$

Can anyone give me few examples which is not defined piece wise as above?

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Before I think about $|\cdot |$, does it count as a piece wise function? –  Git Gud Jan 16 '13 at 18:58
what would that be? could you give me link about |.| –  Santosh Linkha Jan 16 '13 at 18:59
I mean the absolute value. Of what? Yet to be determined, maybe $|x|$ or $|xy|$.. –  Git Gud Jan 16 '13 at 19:00
I think not .. i just don't want that condition imposed like f(x) = this when x = this or x=that –  Santosh Linkha Jan 16 '13 at 19:01
It's probably continuous anyway. –  Git Gud Jan 16 '13 at 19:02
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3 Answers

up vote 1 down vote accepted


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clever manipulation!! –  Santosh Linkha Jan 16 '13 at 19:36
@experimentX And I know you can't disallow limits if you would be willing to accept $e^x$ or $\sin x$. :) –  Hagen von Eitzen Jan 16 '13 at 19:37
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The standard "elementary" functions are always continuous where they are defined, so this would be hard to do. You might try $$ f(x,y) = \arg( -\exp(i(y-x^2)(2x^2-y))) - (y-x^2)(2x^2-y) $$ where arg is the "principal branch" of the argument.

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$f(x,y)=\frac{xy}{x^2+y^2}$ at $(x,y)\neq (0,0)$ and $=0$ at $(x,y)=(0,0)$

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OP asked for the function to not bet piece wise defined. –  Git Gud Jan 16 '13 at 18:55
also I am looking specifically (if possible) not piecewise functions with conditions –  Santosh Linkha Jan 16 '13 at 18:56
Since $f(x,x)\to\frac12$ when $x\to0$, some functional derivatives of $f$ at $(0,0)$ do not exist. –  Did Jan 16 '13 at 19:08
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